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What is a Plant Cell? Plant cells are eukaryotic cells that vary in several fundamental factors from other eukaryotic organisms. Both plant and animal cells contain a nucleus along with similar organelles. One of the distinctive aspects of a plant cell is the presence of a cell wall outside the cell membrane. Plant Cell Structure Just like different organs within the body, plant cell structure includes various components known as cell organelles that perform different functions to sustain itself. These organelles include: Cell Wall It is a rigid layer which is composed of polysaccharides cellulose, pectin and hemicellulose. It is located outside the cell membrane. It also comprises glycoproteins and polymers such as lignin, cutin, or suberin. The primary function of the cell wall is to protect and provide structural support to the cell. The plant cell wall is also involved in protecting the cell against mechanical stress and providing form and structure to the cell. It also filters the molecules passing in and out of it. The formation of the cell wall is guided by microtubules. It consists of three layers, namely, primary, secondary and the middle lamella. The primary cell wall is formed by cellulose laid down by enzymes. Cell membrane It is the semi-permeable membrane that is present within the cell wall. It is composed of a thin layer of protein and fat. The cell membrane plays an important role in regulating the entry and exit of specific substances within the cell. For instance, cell membrane keeps toxins from entering inside, while nutrients and essential minerals are transported across. Nucleus The nucleus is a membrane-bound structure that is present only in eukaryotic cells. The vital function of a nucleus is to store DNA or hereditary information required for cell division, metabolism and growth. 1. Nucleolus: It manufactures cellsâ protein-producing structures and ribosomes. 2. Nucleopore: Nuclear membrane is perforated with holes called nucleopore that allow proteins and nucleic acids to pass through. Plastids They are membrane-bound organelles that have their own DNA. They are necessary to store starch and to carry out the process of photosynthesis. It is also used in the synthesis of many molecules, which form the building blocks of the cell. Some of the vital types of plastids and their functions are stated below: Leucoplasts They are found in the non-photosynthetic tissue of plants. They are used for the storage of protein, lipid and starch. Chromoplasts They are heterogeneous, colored plastid which is responsible for pigment synthesis and for storage in photosynthetic eukaryotic organisms. Chromoplasts have red-, orange- and yellow-colored pigments which provide color to all ripe fruits and flowers. Central Vacuole It occupies around 30% of the cellâs volume in a mature plant cell. Tonoplast is a membrane that surrounds the central vacuole. The vital function of the central vacuole apart from storage is to sustain turgor pressure against the cell wall. The central vacuole consists of cell sap. It is a mixture of salts, enzymes and other substances. Golgi Apparatus They are found in all eukaryotic cells, which are involved in distributing synthesized macromolecules to various parts of the cell. Ribosomes They are the smallest membrane-bound organelles which comprise RNA and protein. They are the sites for protein synthesis, hence, also referred to as the protein factories of the cell. Mitochondria They are the double-membraned organelles found in the cytoplasm of all eukaryotic cells. They provide energy by breaking down carbohydrate and sugar molecules, hence they are also referred to as the âPowerhouse of the cell.â Lysosome Lysosomes are called suicidal bags as they hold digestive enzymes in an enclosed membrane. They perform the function of cellular waste disposal by digesting worn-out organelles, food particles and foreign bodies in the cell. In plants, the role of lysosomes is undertaken by the vacuoles. Chloroplasts It is an elongated organelle enclosed by phospholipid membrane. The chloroplast is shaped like a disc and the stroma is the fluid within the chloroplast that comprises a circular DNA. Each chloroplast contains a green colored pigment called chlorophyll required for the process of photosynthesis. The chlorophyll absorbs light energy from the sun and uses it to transform carbon dioxide and water into glucose. Structure of Chloroplast Chloroplasts are found in all higher plants. It is oval or biconvex, found within the mesophyll of the plant cell. The size of the chloroplast usually varies between 4-6 Âľm in diameter and 1-3 Âľm in thickness. They are double-membrane organelle with the presence of outer, inner and intermembrane space. There are two distinct regions present inside a chloroplast known as the grana and stroma. ⢠Grana are made up of stacks of disc-shaped structures known as thylakoids or lamellae. The granum of the chloroplast consists of chlorophyll pigments and are the functional units of chloroplasts. ⢠Stroma is the homogenous matrix which contains grana and is similar to the cytoplasm in cells in which all the organelles are embedded. Stroma also contains various enzymes, DNA, ribosomes, and other substances. Stroma lamellae function by connecting the stacks of thylakoid sacs or grana. The chloroplast structure consists of the following parts: Membrane Envelope It comprises inner and outer lipid bilayer membranes. The inner membrane separates the stroma from the intermembrane space. Intermembrane Space The space between inner and outer membranes. Thylakoid System (Lamellae) The system is suspended in the stroma. It is a collection of membranous sacs called thylakoids or lamellae. The green colored pigments called chlorophyll are found in the thylakoid membranes. It is the sight for the process of light-dependent reactions of the photosynthesis process. The thylakoids are arranged in stacks known as grana and each granum contains around 10-20 thylakoids. Stroma It is a colorless, alkaline, aqueous, protein-rich fluid present within the inner membrane of the chloroplast present surrounding the grana. Grana Stack of lamellae in plastids is known as grana. These are the sites of conversion of light energy into chemical energy. Chlorophyll It is a green photosynthetic pigment that helps in the process of photosynthesis. Functions of Chloroplast Following are the important chloroplast functions: ⢠The most important function of the chloroplast is to synthesize food by the process of photosynthesis. ⢠Absorbs light energy and converts it into chemical energy. ⢠Chloroplast has a structure called chlorophyll which functions by trapping the solar energy and is used for the synthesis of food in all green plants. ⢠Produces NADPH and molecular oxygen (O 2 ) by photolysis of water. ⢠Produces ATP â Adenosine triphosphate by the process of photosynthesis. ⢠The carbon dioxide (CO2) obtained from the air is used to generate carbon and sugar during the Calvin Cycle or dark reaction of photosynthesis. Mitochondria âMitochondria are membrane-bound organelles present in the cytoplasm of all eukaryotic cells, that produce adenosine triphosphate (ATP), the main energy molecule used by the cell.â What are Mitochondria? Popularly known as the âPowerhouse of the cell,â mitochondria (singular: mitochondrion) are a double membrane-bound organelle found in most eukaryotic organisms. They are found inside the cytoplasm and essentially function as the cellâs âdigestive system.â They play a major role in breaking down nutrients and generating energy-rich molecules for the cell. Many of the biochemical reactions involved in cellular respiration take place within the mitochondria. The term âmitochondrionâ is derived from the Greek words âmitosâ and âchondrionâ which means âthreadâ and âgranules-likeâ, respectively. It was first described by a German pathologist named Richard Altmann in the year 1890. Structure of Mitochondria ⢠The mitochondrion is a double-membraned, rod-shaped structure found in both plant and animal cell. ⢠Its size ranges from 0.5 to 1.0 micrometers in diameter. ⢠The structure comprises an outer membrane, an inner membrane, and a gel-like material called the matrix. ⢠The outer membrane and the inner membrane are made of proteins and phospholipid layers separated by the intermembrane space. ⢠The outer membrane covers the surface of the mitochondrion and has a large number of special proteins known as porins. Cristae The inner membrane of mitochondria is rather complex in structure. It has many folds that form a layered structure called cristae, and this helps in increasing the surface area inside the organelle. The cristae and the proteins of the inner membrane aid in the production of ATP molecules. The inner mitochondrial membrane is strictly permeable only to oxygen and ATP molecules. A number of chemical reactions take place within the inner membrane of mitochondria. Mitochondrial Matrix The mitochondrial matrix is a viscous fluid that contains a mixture of enzymes and proteins. It also comprises ribosomes, inorganic ions, mitochondrial DNA, nucleotide cofactors, and organic molecules. The enzymes present in the matrix play an important role in the synthesis of ATP molecules. Functions of Mitochondria The most important function of mitochondria is to produce energy through the process of oxidative phosphorylation. It is also involved in the following process: 1. Regulates the metabolic activity of the cell 2. Promotes the growth of new cells and cell multiplication 3. Helps in detoxifying ammonia in the liver cells 4. Plays an important role in apoptosis or programmed cell death 5. Responsible for building certain parts of the blood and various hormones like testosterone and estrogen 6. Helps in maintaining an adequate concentration of calcium ions within the compartments of the cell 7. It is also involved in various cellular activities like cellular differentiation, cell signaling, cell senescence, controlling the cell cycle and in cell growth. Disorders Associated with Mitochondria Any irregularity in the way mitochondria function can directly affect human health, but often, it is difficult to identify because symptoms differ from person to person. Disorders of the mitochondria can be quite severe; in some cases, they can even cause an organ to fail.
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: ⢠Free-falling objects do not encounter air resistance. ⢠All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs ⢠Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 ⢠(-8.00 m/s2) ⢠d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) ⢠d (16.0 m/s2) ⢠d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) ⢠(4.1 s) + ½ ⢠(6.00 m/s2) ⢠(4.10 s)2 d = (0 m) + ½ ⢠(6.00 m/s2) ⢠(16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: ⢠An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. ⢠If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) ⢠(t) + ½ ⢠(-9.8 m/s2) ⢠(t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) ⢠(t)2 -8.52 m = (-4.9 m/s2) ⢠(t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 â˘(-9.8m/s2) â˘d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) â˘d (-19.6 m/s2) ⢠d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) ⢠d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
Of the 7 billion people on Earth roughly 0:02 6 billion own a cell phone which is 0:05 pretty shocking given that only 4 and2 0:07 billion have access to a working toilet 0:09 so how are these popular gadgets 0:11 changing your body and brain If you're 0:13 looking down at your phone right now 0:15 your spine angle is equivalent to that 0:17 of an 8-year-old child sitting on your 0:19 neck which is fairly significant 0:21 considering people spend an average of 0:23 4.7 hours a day looking at their phone 0:26 this combined with the length of time 0:28 spent in front of computers has led to 0:30 an increase in the prevalence of myopia 0:32 or nearsightedness in North America in 0:34 the 1970s about one quar of the 0:36 population had myopia where today nearly 0:39 half do and in some parts of Asia 80 to 0:41 90% of the population is now nearsighted 0:44 and it can be hard to put your phone 0:46 down take for example the game Candy 0:48 Crush as you play the game you achieve 0:50 small goals causing your brain to be 0:52 rewarded with little bursts of dopamine 0:54 and eventually you're rewarded in the 0:56 game with new content this novelty also 0:58 gives little bursts of dopamine and 1:00 together create what is known as a 1:01 compulsion Loop which just happens to be 1:04 the same Loop responsible for the 1:05 behaviors associated with nicotine or 1:07 cocaine our brains are hardwired to make 1:10 us novelty seeking and this is why apps 1:12 on our phones are designed to constantly 1:14 provide us with new content making them 1:16 hard to put down as a result 93% of 1:19 young people aged 18 to 29 report using 1:21 their smartphone as a tool to avoid 1:23 boredom as opposed to other activities 1:26 like reading a book or engaging with 1:27 people around them this has created a 1:29 new term nomophobia the fear or anxiety 1:32 of being without your phone we also see 1:35 a change in brain patterns Alpha rhythms 1:37 are commonly associated with wakeful 1:39 relaxation like when your mind wanders 1:41 off whereas gamma waves are associated 1:44 with conscious attentiveness and 1:46 experiments have shown that when a cell 1:47 phone is transmitting say during a phone 1:49 call the power of these Alpha Waves is 1:52 significantly boosted meaning phone 1:54 Transmissions can literally change the 1:56 way your brain functions your smartphone 1:58 can also disrupt your sleep the screen 2:00 emits a blue light which has been shown 2:02 to alter our circadian rhythms 2:03 diminishing the time spent in deep Sleep 2:06 which is linked to the development of 2:07 diabetes cancer and obesity Studies have 2:10 shown that people who read on their 2:11 smartphone at night have a harder time 2:13 falling asleep and produce less 2:15 melatonin a hormone responsible for the 2:17 regulation of sleep wake Cycles Harvard 2:20 medical school advises the last 2 to 3 2:22 hours before bed be technology free so 2:24 pick up a book before bed instead of 2:26 course smartphones also completely 2:28 change our ability to access information 2:30 most notably in poor and minority 2:32 populations 7% of Americans are entirely 2:35 dependent on smartphones for their 2:37 access to the internet a 2014 study 2:40 found that the majority of smartphone 2:41 owners use their phone for online 2:43 banking to look up medical information 2:45 and searching for jobs so while phones 2:47 are in no way exclusively bad and have 2:50 been part of a positive change in the 2:51 world there's no denying that they are 2:53 changing us but many successful people 2:56 have now decided to take smartphone 2:58 vacations in order to increase 3:00 productivity in our new ASAP thought 3:01 video we break down the top six reasons 3:04 you should take a smartphone vacation 3:06 and how it could benefit your life right 3:08 now and subscribe for more weekly 3:09 science videos
Short Quiz in Math 1 visualizes and represents numbers from 0 to 100 using a variety of materials
Make a multiple choice quiz for my year 8 science students based on the science in this transcript from a video: 3°C 0:04 It can be the difference between snow and sleet 0:08 Wearing a jacket or not 0:11 In your day-to-day life, it may not seem significant 0:15 But 3°C of global warming would be catastrophic 0:20 Heatwaves, droughts, extreme precipitation, even fire 0:25 3°C of warming is really disastrous 0:28 The scary thing is, the world is well on its way there 0:32 Since the industrial revolution, the Earth has warmed between 1.1°C and 1.3°C 0:40 This is a problem that babies you pass in the street will have to live with 0:46 Children born today... 0:47 ...are up to seven times more likely to face extreme weather than their grandparents 0:52 If global temperatures do rise by 3°C... 0:55 ...what would their world look like? Climate change is already having devastating effects 1:03 Rising sea levels 1:05 Desertification 1:07 Hollywood has always enjoyed imagining the end of the world 1:11 While blockbusters like this are clearly fiction... 1:14 ...this film will show the scenario we all face... 1:17 ...unless more drastic measures are taken to stop burning fossil fuels 1:30 In some parts of the world the effects of inaction are already clear 1:35 The slums of Bangladeshâs capital are filling up with climate migrants 1:41 Minara comes from Bhola District, an area in southern Bangladesh 1:46 There, like many other parts of the country... 1:49 ...rivers swollen by heavier rain and melting Himalayan glaciers... 1:53 ...are washing away peopleâs homes 1:56 Many, like her, have lost everything 2:00 Our home in Bhola had endless amounts of land 2:03 There was lots of space for farming, we had a spacious house 2:08 There were different types of fruits, vegetation and trees growing at home 2:12 We used to eat the fruit from our own trees 2:18 I canât eat them now because they don't exist anymore 2:21 Since the river flooded for the third time, I had to flee to Dhaka 2:26 Life was much better back home 2:29 It was unbearable to live through, truly intolerable 2:33 We didnât have the time to save anything at all 2:38 1.1°C to 1.3°C of global warming has already transformed Minaraâs life 2:45 Itâs one of the reasons why so many migrants like her... 2:47 ...are moving to the city each year... 2:50 ...nearly 400,000 according to the last estimate 2:53 And climate models show there could be much worse to come How climate modelling works 3:02 Climate scientist Joeri Rogelj... 3:04 ...has spent the last ten years modelling future climate scenarios... 3:08 ...for the United Nations 3:10 The models we use to carry out this exercise... 3:13 ...really represent the state of the art... 3:15 ...of our current knowledge of climate change and where we are heading 3:19 Joeriâs projections use data collected by hundreds of scientists around the world 3:26 Here this is the 3°C level... 3:28 ...and so there is at least a one-in-four chance that under current policies... 3:32 ...we would hit 3°C by the end of the century 3:36 This is just one of the scenarios Joeri looks at 3:40 Another one imagines that all policy promises are kept 3:44 The most optimistic assumes that all promises have been kept... 3:47 ...and net-zero targets are met 3:50 Where our best estimate ends up around 2°C at the end of the century... 3:54 ...there is still a one-in-20 chance that we end up with 3°C instead 3:59 One would not be entering a plane if there is a one-in-20 chance... 4:03 ...that the plane will crash Nowhere is safe from global warming 4:07 A rise of 3°C would affect everyone 4:10 Even wealthy cities in rich countries wouldnât be immune to the consequences 4:15 European capitals like Paris and Berlin... 4:18 ...would bake under more extreme heatwaves 4:22 Frequent storm-surges in New York could turn parts of the city desolate 4:27 In many ways, cities magnify, intensify climate events 4:33 Cities are hotter than the places around them... 4:36 ...they tend to be more vulnerable to flooding 4:39 And you can get a really bad event in a city in a way that you canât in the countryside 4:46 And because of their denser populations... 4:49 ...disasters in a city affect far more people 4:52 Some cities might be badly prepared for the changes coming 4:56 But they have the means to adapt 4:59 Cities tend to be wealthier than surrounding places 5:03 They have a lot of amenities 5:05 A city that has taken seriously the risks of a 3°C world... 5:08 âŚwouldnât necessarily be a worse place to be in a 3°C world 5:12 But a city that hasnât prepared for these sort of eventualities... 5:16 ...that might be a really nasty place The impact of prolonged droughts 5:20 So far, many developed cities have got off lightly... 5:24 ...but some rural parts of the world are suffering disproportionately 5:29 Smallholdersâsmall-scale farmersâare particularly vulnerable to climate change 5:35 And there are over 600 million around the world 5:38 Smallholders with farms under two hectares... 5:40 ...produce around a third of the global food supply 5:46 Central Americaâs âDry Corridorâ... 5:48 ...supports a mix of smallholdings and medium-sized farms 5:53 Sandwiched between the Pacific Ocean and the Caribbean Sea... 5:56 ...the area is prone to droughts 6:08 Israel RamĂrez Rivera is a smallholder in Guatemala 6:12 Here, climate change is making the dry seasons longer, and more severe 6:18 This is the biggest ear of maize that this plot could deliver 6:23 He depends on his crops of corn and beans 6:26 But theyâre getting harder to grow 6:30 The surrounding mountains... 6:32 ...used to provide us with native food... 6:38 ...and now that isnât an option anymore... 6:41 ...due to climate change and its effects 6:46 Nearly two-thirds of the smallholders in the Dry Corridor now live in poverty 6:52 The impact of all of this for us... 6:59 ...malnutrition among children 7:03 Weâve lost a few 7:07 For my crops especially, the midsummer heat is harder than before 7:16 The plant dries up and canât provide us... 7:19 ...with the necessary food provision 7:24 Severe droughts in Central America... 7:26 ...are now four times more likely than they were last century 7:30 Many families from here have gone to the States 7:37 The economic despair and debts... 7:44 ...have pushed many people from this community to do this journey 7:53 Migration from Guatemala to the United States has quadrupled since 1990 7:59 Not all of this has been due to climate change 8:02 But longer droughts would force even more to move 8:05 In a 3°C world, annual rainfall in this region... 8:09 ...could drop by up to 14% 8:12 At 3°C, over a quarter of the worldâs population... 8:16 ...could endure extreme droughts for at least a month of the year 8:19 Northern Africa could see droughts that last for years at a time Rising sea levels, storm surges and flooding 8:24 But for some, too much water will be the problem 8:29 10% of the worldâs population lives on a coastline... 8:32 ...thatâs less than 10 metres above sea level 8:35 For these coastal inhabitants, a 3°C world would spell disaster 8:40 By 2100, global sea levels could have climbed by half a metre from 2005 levels 8:46 Low-lying cities like Lagos would be especially vulnerable... 8:49 ...with up to up to a third of the population displaced 8:54 And in Fiji, rising waters are already upending lives 9:04 You can see the graveyard there, itâs all under water now... 9:08 ...due to this rising sea level and climate change 9:15 The village of Togoru in Fiji is being swallowed by the sea 9:19 Barney Dunn, the village headman, has seen over half the village disappear 9:24 Relativesâ houses have been abandoned, and family graves are now under water 9:29 We have been asked by the government to relocate... 9:32 ...but no one wants to relocate... 9:34 ...because we have our great-great-grandparents down there in the sea 9:39 This is the place weâve been brought up in 9:41 ...itâs not easy to leave 9:44 Past attempts to build a seawall havenât worked 9:48 But Barney sees building a new one as the villageâs only hope 9:52 If they do that, maybe we can save whatever is left 9:56 But if we donât have the seawall, then it will be keep eroding and time will come... 10:01 ...maybe in ten,15 years, Togoru will be all eroded 10:05 Rising seas also mean storms cause more floods 10:11 And many more countries could suffer 10:14 The Philippines and Myanmar are just two countries... 10:17 ...that will also see an increase in storm surges in a 3°C world 10:21 To escape, many will move⌠10:24 âŚoften, to urban areas Extreme heat and wet-bulb temperatures 10:27 Half the worldâs population already lives in cities... 10:31 ...almost a third in slums 10:36 For them, a 3°C world could be deadly 10:40 Minara has moved to Dhaka to escape the impact of climate change 10:44 But life could get even worse for her 10:47 Iâm struggling a lot nowadays 10:49 The heat during the day is unbearable 10:52 Even late at night it doesnât cool down 10:57 The heat is getting more intense every day 10:59 I mean, itâs going to get much worse 11:03 I can barely survive it now, how will I live through it in the future? 11:08 Dhaka is getting hotter 11:11 In the last 20 years the average daytime temperature... 11:13 ...has crept up by nearly half a degree 11:17 Days that approach 40°C are now being reported 11:20 And high so-called wet-bulb temperatures are on the rise 11:26 A wet-bulb temperature is a measure of heat and humidity 11:30 Humans cool themselves by sweating⌠11:32 But in these conditions, when relative humidity is near 100%... 11:36 ...sweat doesnât evaporate well 11:38 So people canât cool down⌠11:41 ...even if given unlimited shade and water 11:45 At a high wet-bulb temperature, the body canât lose heat... 11:49 ...and so it gets hotter and hotter... 11:51 ...and the body is designed to work at a given temperature 11:53 And if it gets too hot inside, you will die 11:58 The human limit for wet-bulb temperatures is 35°C... 12:02 ...around skin temperature 12:04 Dhaka will have a much higher chance... 12:05 ...of reaching dangerous wet-bulb temperatures... 12:07 ...if global warming reaches 3°C 12:12 You canât really adapt to that 12:14 You have to get out. If the temperature is so high that you canât work... 12:20 ...canât do hard manual labour outside for significant parts of the year... 12:25 ...then many places will become functionally no longer part of the economy 12:33 Jacobabad in Pakistan, and Ras al Khaimah, in the United Arab Emirates... 12:37 ...have already recorded deadly wet-bulb temperatures 12:40 More of the tropics and the Persian Gulf... 12:43 ...as well as parts of Mexico and the south-eastern United States... 12:47 ...could all get to this threshold by the end of the century 12:50 Climate modelling might show us the weather Increased migration and conflict 12:52 But it doesnât show us its other effects on society 12:56 Established migration patterns could change 12:59 Climate disasters may exacerbate reasons people cross borders 13:03 Within countries, more people will move to cities 13:07 In a 3°C world, tens of millions of people a year... 13:10 ...could be displaced by disasters made worse by climate change 13:15 When people are displaced by climate... 13:18 âŚthey may well go to cities... 13:19 ...because cities are the places that attract people from the countryside already 13:25 A lot of people who can get to the developed world... 13:28 ...not least because the developed world tends to be less hot, will give that a go 13:35 As migration around the world increases... 13:38 ...there could be more competition for fewer resources 13:42 Waterâalready a highly contested resourceâwill be a focal point 13:47 Turkeyâs new Ilisu dam has reduced the flow of water into Iraq 13:53 China lays claim to rivers vital to India and Pakistan 13:57 The prospect of a water-conflict makes people very uneasy 14:03 How national tensions would exacerbate those sorts of reactions... 14:08 ...in a 3°C world... 14:09 ...is the sort of thing that no one should really want to find out 14:14 I think youâd have to be incredibly sanguine... 14:16 ...not to think that the sort of climate extremes that we talk about... 14:19 ...in a 3°C world wouldnât lead some places... 14:22 ...to the brink of societal collapse 14:25 Those lucky enough to escape unrest... Adaptation and mitigation are crucial 14:28 ...would still have to adapt to a radically different world 14:32 People can adapt to climate change in all sorts of ways, one of the most obvious ones... 14:37 ...is air conditioning 14:39 But other ways to adapt at a local or regional level... 14:42 ...I mean, one of the most obvious is diversifying agriculture 14:47 There are physical things you can do, like seawalls 14:52 The fact that people can adapt and that adaptation will reduce suffering... 14:57 ...doesnât mean that it will eliminate suffering 15:00 Suffering is built into this whole process of heating up the planet 15:06 Adaptation will only get the world so far 15:09 The best way to deal with a 3°C world... 15:12 ...is not to go to a 3°C world 15:14 And thatâs why increasing efforts on mitigation are important 15:17 Itâs why working towards negative emissions... 15:20 ...that could bring down the temperature after it peaks are important 15:25 Once you get to a 3°C world, you are in real bad global trouble 15:33 The scale of change needed... 15:35 ...and the slow progress of governments so far... 15:38 ...means 3°C of warming is uncomfortably likely unless more is done 15:44 Despite existing pledges, greenhouse-gas emissions... 15:48 ...are still set to rise by 16% from 2010 levels by 2030 15:54 The need to act has never been clearer 15:57 Thereâs still time to reduce emissions, so that a 3°C world remains fiction... 16:02 ...rather than becoming fact
Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, âŚ. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the SchrĂśdinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, âŚ, n â 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction Ď is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n â l â 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 â 0 â 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 â 0 â 1 = 1 node for 2s, and 3 â 0 â 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l â 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to â1, 0, or +1. Generally speaking, ml can be equal to âl, â(l â 1), âŚ, â1, 0, +1, âŚ, (l â 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ⼠1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electronâelectron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the SchrĂśdinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the SchrĂśdinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron ârotationâ or âspinning.â Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only âspinâ in one of two quantized states. One is termed the Îą state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=â12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=â12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of â12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=â12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (Âą12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly
A solution is a mixture in which one or more substances are uniformly distributed in another substance. Solutions can be mixtures of liquids, solids, or gases. For example, plasma, the liquid part of blood, is a very complex solution. It is composed of many types of ions and large molecules, as well as gases, that are dissolved in water. A solute (SAHL-YOOT) is a substance dissolved in the solvent. The particles that compose a solute may be ions, atoms, or molecules. The solvent is the substance in which the solute is dissolved. For example, when sugar, a solute, and water, a solvent, are mixed, a solution of sugar water results. Though the sugar dissolves in the water, neither the sugar molecules nor the water molecules are altered chemically. If the water is boiled away, the sugar molecules remain and are unchanged. Solutions can be composed of various proportions of a given solute in a given solvent. Thus, solutions can vary in concentra- tion. The concentration of a solution is the amount of solute dis- solved in a fixed amount of the solution. For example, a 2 percent saltwater solution contains 2 g of salt dissolved in enough water to make 100 mL of solution. The more solute dissolved, the greater is the concentration of the solution. A saturated solution is one in which no more solute can dissolve. Aqueous (AY-kwee-uhs) solutionsâsolutions in which water is the solventâare universally important to living things. Marine microorganisms spend their lives immersed in the sea, an aqueous solution. Most nutrients that plants need are in aqueous solutions in moist soil. Body cells exist in an aqueous solution of intercellu- lar fluid and are themselves filled with fluid; in fact, most chemical reactions that occur in the body occur in aqueous solutions. Copyright Š by Holt, Rinehart and Winston. All rights reserved. Liquid water Solid water Ice (solid water) is less dense than liquid water because of the structure of ice crystals. The water molecules in ice are bonded to each other in a way that creates large amounts of open space between the molecules, relative to liquid water. FIGURE 2-12 solvent from the Latin solvere, meaning âto loosenâ Word Roots and Origins CHEMISTRY OF LIFE 43 ACIDS AND BASES One of the most important aspects of a living system is the degree of its acidity or alkalinity. What do we mean when we use the terms acid and base? Ionization of Water As water molecules move about, they bump into one another. Some of these collisions are strong enough to result in a chemical change: one water molecule loses a proton (a hydrogen nucleus), and the other gains this proton. This reaction really occurs in two steps. First, one molecule of water pulls apart another water molecule, or dissociates, into two ions of opposite charge: H2O â H OH The OH ion is known as the hydroxide ion. The free H ion can react with another water molecule, as shown in the equation below. H H2O â H3O The H3O ion is known as the hydronium ion. Acidity or alkalin- ity is a measure of the relative amounts of hydronium ions and hydroxide ions dissolved in a solution. If the number of hydronium ions in a solution equals the number of hydroxide ions, the solution is said to be neutral. Pure water contains equal numbers of hydro- nium ions and hydroxide ions and is therefore a neutral solution. Acids If the number of hydronium ions in a solution is greater than the number of hydroxide ions, the solution is an acid. For example, when hydrogen chloride gas, HCl, is dissolved in water, its mol- ecules dissociate to form hydrogen ions, H, and chloride ions, Cl, as is shown in the equation below. HCl â H Cl These free hydrogen ions combine with water molecules to form hydronium ions, H3O. This aqueous solution contains many more hydronium ions than it does hydroxide ions, making it an acidic solution. Acids tend to have a sour taste; how- ever, never taste a substance to test it for acidity. In concentrated forms, they are highly corrosive to some materials, as you can see in Figure 2-13. Bases If sodium hydroxide, NaOH, a solid, is dissolved in water, it dissociates to form sodium ions, Na, and hydroxide ions, OH, as shown in the equation below. NaOH â Na OH Copyright Š by Holt, Rinehart and Winston. All rights reserved. Eco Connection onnection Acid Precipitation Acid precipitation, more commonly called acid rain, describes rain, snow, sleet, or fog that contains high levels of sulfuric and nitric acids. These acids form when sulfur dioxide gas, SO2, and nitrogen oxide gas, NO, react with water in the atmosphere to produce sulfuric acid, H2SO4, and nitric acid, HNO3. Acid precipitation makes soil and bodies of water, such as lakes, more acidic than normal. These high acid levels can harm plant and animal life directly. A high level of acid in a lake may kill mollusks, fish, and amphibians. Even in a lake that does not have a very elevated level of acid, acid precipitation may leach aluminum and magnesium from soils, poisoning water- dwelling species. Reducing fossil-fuel consump- tion, such as occurs in gasoline engines and coal-burning power plants, should reduce high acid levels in precipitation. Sulfur dioxide, SO2, which is produced when fossil fuels are burned, reacts with water in the atmosphere to produce acid precipitation. Acid precipitation, or acid rain, can make lakes and rivers too acidic to support life and can even corrode stone, such as the face of this statue. FIGURE 2-13 44 CHAPTER 2 This solution then contains more hydroxide ions than hydronium ions and is therefore defined as a base. The adjective alkaline refers to bases. Bases have a bitter taste; however, never taste a substance to test for alkalinity. They tend to feel slippery because the OH ions react with the oil on our skin to form a soap. In fact, commercial soap is the product of a reaction between a base and a fat. pH Scientists have developed a scale for comparing the relative con- centrations of hydronium ions and hydroxide ions in a solution. This scale is called the pH scale, and it ranges from 0 to 14, as shown in Figure 2-14. A solution with a pH of 0 is very acidic, a solution with a pH of 7 is neutral, and a solution with a pH of 14 is very basic. A solutionâs pH is measured on a logarithmic scale. That is, the change of one pH unit reflects a 10-fold change in the acidity or alkalinity. For example, urine has 10 times the H3O ions at a pH of 6 than water does at a pH of 7. Vinegar, has 1,000 times more H3O ions at a pH of 3 than urine at a pH of 6, and 10,000 times more H3O ions than water at a pH of 7. The pH of a solution can be measured with litmus paper or with some other chemical indicator that changes color at various pH levels. Buffers The control of pH is important for living systems. Enzymes can function only within a very narrow pH range. The control of pH in organisms is often accomplished with buffers. Buffers are chemi- cal substances that neutralize small amounts of either an acid or a base added to a solution. As Figure 2-14 shows, the composition of your internal environmentâin terms of acidity and alkalinityâ varies greatly. Some of your body fluids, such as stomach acid and urine, are acidic. Others, such as intestinal fluid and blood, are