Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 301:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a

  Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  The mass in the first experiment is pulled down a distance A from its equilibrium position and then released from rest. The mass will then oscillate with simple harmonic motion. As the mass moves up and down, energy is dissipated due to factors such as air resistance and internal heating of the spring. The mass will no longer oscillate when the total energy dissipated equals:

  A. kL2/2

  B. kA2/2

  C. k(L + A)2/2

  D. kL0 2/2

  Correct Answer: B

• Question 302:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a

  Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  In the first experiment the mass is pulled down and set into motion. The position of greatest speed is:

  A. at the equilibrium position.

  B. at the position where the spring's length is its natural length.

  C. at the lowest point in its motion.

  D. at the highest point in its motion.

  Correct Answer: A

  First note that the motion of the spring-mass system is simple harmonic motion. When simple harmonic motion occurs, the speed is a maximum at the equilibrium position, so answer choice A is the correct answer. For a horizontal spring-mass system the equilibrium position of the mass corresponds to the spring being neither stretched nor compressed. But as the passage explains, with a vertical spring-mass system the situation is different. When the mass is in equilibrium, the spring is stretched--and that makes choice B incorrect. As for choices C and D, they represent the extreme positions in the motion of the mass. We are told at the beginning of the passage that the mass is moving with simple harmonic motion, so in these positions the displacement from the equilibrium position is a maximum and the speed is zero. If you picked choice C, the lowest point, you might have been thinking of something being dropped from a height h, but that is not what is happening here.

• Question 303:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at –20°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  What would be the pH of 100 mL of the sodium acetate stock solution after the addition of 3.6 g of HCl? (pKa of acetic acid = 4.74)

  A. 1.0

  B. 4.74

  C. 5.2

  D. 6.0

  Correct Answer: B

  In an aqueous solution, sodium acetate is in equilibrium with its corresponding acid:

  

  In this equilibrium, acetic acid acts as a weak acid and acetate acts as a conjugate base; this constitutes a buffer system. Buffer systems resist large changes in pH upon the addition of small amounts of acid or base. For example, when a base is added to the acetic acid/acetate buffer, the hydroxide ions are consumed by acetic acid to form acetate and water, so the pH doesn't rise too drastically. When an acid is added to the buffer, the hydrogen ions are consumed by acetate to form the undissociated acid, so the pH doesn't drop too drastically. In order to work out the change in pH, we have to first calculate the initial concentrations of acetic acid and sodium acetate. We already know from Table 1 that the concentration of sodium acetate is 3M. Plugging this into the Henderson-Hasselbalch equation, we can work out the concentration of acetic acid. The Henderson-Hasselbalch equation is as follows: pH = pKa + log [A–]/[HA], where [A–] is the acid concentration of the conjugate base, and [HA] is the concentration of the acid. Plugging our values into his equation, we get 5.2 = 4.74 + log [3]/[HA]. The concentration of acetic acid, therefore, works out to be 1M. In a 100mL solution, the number of moles of sodium acetate and acetic acid works out to be 0.3 and 0.1 moles, respectively. Now, what is the pH of the solution after 3.6g of hydrochloric acid is added? The molar mass of hydrochloric acid is 36.5g/mol, so the number of moles is 3.6/36.5 or 0.1. Remember, when an acid is added to the buffer, the hydrogen ions are consumed by acetate to form undissociated acetic acid. Therefore, if there are 0.3 moles of sodium acetate in the original solution, there will be 0.3 – 0.1 moles after the addition of hydrochloric acid. Similarly, if there are 0.1 moles of acetic acid in the original solution, there will be 0.1 + 0.1 moles after the addition of HC1. Therefore, the new concentration of acetic acid and acetate is 2M. Plugging these new values into the Henderson-Hasselbach equation we get pH = 4.74 + log [2]/[2]. Since the log of 1 is 0, the pH is equal to the pKa, 4.74. Choice B is, therefore, the correct response. Choice C and choice D can be discarded because we are adding an acid to the buffer system. The only way we can increase the pH of the stock solution is by the addition of base, since the equilibrium we talked about would then shift to the right to form acetate. Since a buffer resists drastic changes in pH, choice A is wrong.

• Question 304:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at –20°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  Which of the following conclusions can be reached based on the fact that DNA precipitates in the last step of the plasmid prep procedure?

  A. DNA dissolves better in water at lower temperatures.

  B. DNA is polar and therefore dissolves better in water than in a mixture of water and ethanol.

  C. DNA is nonpolar and therefore dissolves better in ethanol than in water.

  D. DNA dissolves well in ethanol and precipitates only because the solution is centrifuged.

  Correct Answer: B

  In the last step of the plasmid prep procedure, ethanol is added to the mixture, which is an aqueous solution, or a solution whose solvent is water. Then the test tube is incubated at low temperature for an hour, and finally the solution is centrifuged and a DNA precipitate forms. This happens because DNA, or deoxyribonucleic acid, is a highly polar substance, and is therefore more soluble in an aqueous solution than in a solvent composed mostly of ethanol. Thus, choice B is correct. Remember the rule that like dissolves like: thus a highly polar substance will dissolve better in a more polar solvent -- water -- than in a less polar solvent -- ethanol. Choice c is wrong because there's no evidence in the passage that DNA dissolves better in ethanol than in water; in fact, there's evidence against this conclusion, since the DNA is fully dissolved in the water solution but precipitates out of the water-ethanol solution. Choice D is wrong because centrifugation can't make a substance precipitate out of a solution; it can only make a precipitate that is suspended in the solution settle to the bottom. Finally, choice A is wrong because there's no evidence that DNA dissolves better at lower temperatures; on the contrary, the incubation at ?0?apparently contributes to its precipitation.

• Question 305:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at –20°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  Pure ethanol (CH3CH2OH) is difficult to prepare and therefore expensive; 95% ethanol is much cheaper. Consequently, 95% ethanol is generally used in the preparation of dilute ethanol solutions. How much 95% ethanol would be needed to produce a 500 mL solution of 70% ethanol by volume in water?

  A. 333 mL

  B. 350 mL

  C. 368 mL

  D. 475 mL

  Correct Answer: C

  The easiest way to do it is to figure out how much ethanol is in the solution at the end of the dilution, and work backwards from the end. Now, notice that the meaning of the percentage sign in this question and in the preceding question is different -- there we were talking about percentage by mass, whereas this question tells you clearly that it's dealing with percentage by volume. Also, you're told in the passage that a 70% ethanol solution contains 70% ethanol and 30% water by volume. In practice, this is how researchers do usually measure solutions, just for convenience-- it's easiest to measure liquids by their volume and measure solids by their mass. Anyway, 500 milliliters of a 70% ethanol solution will contain 350 milliliters of ethanol and 150 milliliters of water. So we need an amount of 95% ethanol that contains 350 milliliters of ethanol. If we call this amount X, then 0.95X equals 350 milliliters. So we want a choice that's just a little more than 350, and the only one that fits is C. Sure enough, if we solve the equation, we find that X equals 368 milliliters, choice C.

• Question 306:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at –20°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  What is the molality of a stock solution that is 10% SDS by mass?

  A. 0.028 m

  B. 0.100 m

  C. 0.347 m

  D. 0.385 m

  Correct Answer: D

  To answer this, you have to know the definitions of two measures of solution concentration; percentage by mass and molality. The percentage by mass of a solution is the mass of the solute divided by the total mass of the solution, multiplied by 100. Thus, a 10% SDS solution has 10% of its mass in the form of sodium dodecyl sulfate and 90% of its mass in the form of water. The molality of a solution is the number of moles of solute per kilogram of solvent. Let's assume that this solution contains one kilogram of water. Remember that the water makes up only 90% of the total weight of the solution. We can find the total weight of the solution by using the formula 1000 equals zero point nine x. This tells us that the total weight of the solution is one thousand, one hundred eleven grams, so the weight of the SDS must be one hundred eleven grams. Now we need to know how many moles of SDS there are in the 111 grams. The number of moles is equal to the mass of the SDS over its molecular weight; this comes to 0.38 moles. Since there are 0.385 moles of SDS for every thousand grams of water, the solution is 0.385 molal, choice D.

• Question 307:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at ?0°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  EDTA is available commercially in the form of a hydrated sodium salt, Na2 EDTA ?2H2O. How much of this salt must be used to produce 1 L of a 0.5 M stock solution?

  A. 145 g

  B. 146 g

  C. 186 g

  D. 187 g

  Correct Answer: C

  The first thing we have to do here is figure out the molecular weight of the compound being dissolved, which is Na2 EDTA ?2H2O, or the dihydrate of the disodium salt of ethylene diamine tetraacetic acid. Right there is the trick in this question -- you have to realize that we're not talking about exactly the same compound as the one in the table, but another closely related species. This species is a salt consisting of one molecule of EDTA, plus two molecules of water, plus two atoms of sodium. The two sodium atoms will replace hydrogen atoms in two of the carboxyl groups of EDTA, so you have to subtract the weight of two hydrogen atoms from the weight of the EDTA molecule. You also have to add the weight of two sodiums and of two molecules of water. If we add these all up, we get a total of 372 as the molecular weight of the EDTA salt. To make one liter of a point five molal solution of EDTA, we need half a mole of this salt, which is 186 grams, choice C.

• Question 308:

  A researcher in a molecular biology lab planned to carry out an extraction procedure known as an alkaline plasmid prep, which is designed to purify plasmids, small pieces of the hereditary material DNA, from bacterial cells. The bacteria are first placed into a test tube containing liquid nutrient medium and allowed to grow until they reach a high population density. The culture, which consists of solid cells suspended in the medium, is then centrifuged; a solid pellet is formed. The supernatant is poured out, leaving the pellet behind, and the cells are resuspended in a mL of lysis buffer solution (50 mM glucose, 25 mM Tris buffer and 10 mM ethylenediaminetetraacetic acid (EDTA), with 5 mg of the enzyme lysozyme added). They are then incubated for 30 minutes at 0°C, during which time the bacterial cell walls break down and the cell contents are released into the solution. After incubation, 1 mL of 0.4 N sodium hydroxide and 1 mL of 2% sodium dodecyl sulfate (SDS) are added, and the solution is again incubated on ice for 10 minutes. 2 mL of 3 M sodium acetate are added and the mixture is incubated for 30 minutes at 0°C. The test tube is centrifuged once more and the supernatant is decanted into a clean tube, leaving behind the protein and most other cell components in the pellet. Finally, 10 mL of pure ethanol are added to the supernatant from the previous step to precipitate out the DNA, and the test tube is incubated at ?0°C for 60 minutes, during which the mixture remains liquid. The mixture is centrifuged a final time and the supernatant removed. The translucent precipitate that results is washed with 70% ethanol (70% ethanol and 30% water by volume), allowed to dry, and resuspended in 1 mL of TE buffer (10 mM Tris, 1 mM EDTA). In preparation for this experiment, the researcher prepared stock solutions of the various chemicals that she will need in the experiment. Stock solutions are highly concentrated solutions of commonly used chemicals in water from which dilute solutions are prepared for daily use. Table 1 shows the chemicals, their molecular formulas and weights, and the composition of commonly used stock solutions.

  

  Tris (Tris(hydroxymethyl)aminomethane) is generally used as a buffer. If pH 8.0 is a good buffering region for Tris, then:

  I) the pKa of Tris must be near pH 8.0

  II) if Tris is titrated with acid, the titration curve will possess a steep region near pH 8.0.

  III) a great deal of NaOH would have to be added to pH 8.0 Tris in order to significantly affect the pH.

  A. I only

  B. III only

  C. I and II only

  D. I and III only

  Correct Answer: D

  A buffer is a mixture of either a weak base and its conjugate acid, or a weak acid and its conjugate base. Tris, as it happens, is a base. Buffer solutions resist changes in pH when acid or base is added. If pH 8 is a good buffering region for Tris, the ratio of base (Tris) to its conjugate acid (protonated Tris) will be near to 1. The pH of this solution is equal to the pKa + log [base]/[conjugate acid]. Therefore, if the ratio of base to its conjugate acid is near to 1, the pKa must be near to the pH. Thus Roman numeral I is a true statement. To evaluate Roman numeral II, we have to recall what a titration curve looks like. Titration is a procedure used for determining the normality of an acid or base. The procedure consists of adding an acid to a base, or a base to an acid, until the pH of the mixture reaches 7. The titration curve is a plot showing the pH of the solution as a function of the amount of acid or base added. Since the pH, as the dependent variable, is plotted on the y axis, a steep part of a titration curve represents a rapid change in pH. But we just said that a buffer solution in its effective pH range, such as a pH 8 Tris solution, resets pH change, so the titration curve for Tris will actually be quite flat near pH 8, and so Roman numeral II is false. Adding sodium hydroxide, which is a base, to a pH 8.0 Tris solution also wouldn't change the pH easily, so a large amount of the base would have to be added to affect the pH very much, and Roman numeral III is true. Since I and III are true, the correct answer choice is D.

• Question 309:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are high-energy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  In an X-ray tube, electrons of charge e are accelerated through a potential difference of V. The anode is cooled by water of mass m with specific heat c. If n electrons per second strike the anode, what is the maximum possible rise in the temperature of the water after 100 s?

  A. nVe/100mc

  B. 100Ve/mc

  C. 100n(Ve + mc)

  D. 100nVe/mc

  Correct Answer: D

  We must first calculate the maximum energy that the electrons could transfer to the water. Now, the equation that relates the energy of the electrons E to the total charge q and the potential difference V is E = qV. There are n electrons hitting the anode target per second, therefore 100n electrons strike the target in 100 seconds, which is equal to a charge 1 of 100ne. Therefore, the amount of energy transferred in 100 s = 100ne V. Assuming that all of this energy is converted to heat energy, we can use the equation for specific heat to calculate the change in temperature of the anode. This equation is E = mcT, where E is the energy, m is the mass, c is the specific heat, and T is the change in temperature. This is equal to the energy transferred to the anode target. Setting the two expressions equal to one another, we get that 100nVe = mcT. The question asks us to determine the change in temperature, so if we rearrange this equation to solve for T, we get that the change in temperature T = 100neV/mc, choice D.

• Question 310:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are highenergy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  Which of the following graphs best represents the relationship between the amount of X-rays absorbed per unit length of material and the energy of the X-rays, for lead, bone, and air?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  In this question we are asked to choose which of the four graphs best represents the absorption of X-rays versus the energy of X-rays for three materials with different densities. Now, we are told in the passage that lower energy X-rays are absorbed more readily than higher energy X-rays. Therefore, we are looking for a graph in which the absorption increases as the energy decreases, and this allows us to eliminate answer choices A and D, leaving us with choices B and C. We are also told in the passage that materials with high densities absorb X-rays more readily than materials with lower densities. Recall that lead has a higher density than air, and that bone is intermediate between the two. Therefore, we are looking for a graph that shows the materials in the following order for a particular energy; lead with highest absorption, then bone, then air. The only answer choice that fits this is B, and this is indeed the correct answer.