AP Physics 2: 7.3 Energy in Modern Physics (Energy in Radioactive Decay and  E = mc2) – Exam Style questions with Answer- FRQ

Question: (10 points, suggested time 20 minutes)

Light and matter can be modeled as waves or as particles. Some phenomena can be explained using the wave model, and others can be explained using the particle model.
(a) Calculate the speed, in m/s, of an electron that has a wavelength of 5.0 nm.

(b) The electron is moving with the speed calculated in part (a) when it collides with a positron that is at rest. A positron is a particle identical to an electron except that its charge is positive. The two particles annihilate each other, producing photons. Calculate the total energy of the photons.

(c) A photon approaches an electron at rest, as shown above on the left, and collides elastically with the electron. After the collision, the electron moves toward the top of the page and to the right, as shown above on the right, at a known speed and angle. In a coherent, paragraph-length response, indicate a possible direction for the photon that exists after the collision and its frequency compared to that of the original photon. Describe the application of physics principles that can be used to determine the direction of motion and frequency of the photon that exists after the collision.

▶️Answer/Explanation

Ans:

\(\lambda =\frac{h}{p}\)       5 × 10-9 m = \(\frac{6.63\times 10^{-34}J.5}{(9.11\times 10^{-31}kg)V}\)       p = mv

4.555× 10-39   V = 6.63 × 10-34   

V = 14554.3  m/s

(b)

E = Δmc2      Δm = 2 (9.11 × 10-31 )  = 1.822 × 10-34   

Po = Pf

14554 (me) = mv

E= 1.822 × 10-30 (3×108)2

E = 1.64 × 10-13 J

(c)

The initial momentum has to equal to final momentum in all  directions. The frequency of the original will be greater than the final frequency as energy must be conserved and it depends on frequency. The photon will travel back towards the bottom of the page as to conserve momentum of the vertical axis but the angle will be less than the election to conserve horizontal momentum.

Question

Given the following information:
                                            Proton mass = 1 .0078 u
                                            Neutron mass = 1 .0087 u
                                            Mass of \(^{226}_{88}Ra\) = 226.0244 u
(a) Determine the mass defect for this isotope of radium.
(b) What does this mass defect represent? Explain both qualitatively and quantitatively.
(c) If radium-88 naturally undergoes alpha decay, write down a nuclear reaction for this process. Be sure to show any energy required ( Q ) or released by this process. If a new element is formed and you are unsure of its symbol, you may use an X to represent that new element. Use the same isotope notation as that given in the information above.
(d) In the reaction in part (c), compare the total mass defects on the reactant side to the mass defect found in part (a):
_____ The reactants have the same mass defect.
_____ The reactants have a larger mass defect.
_____ The reactants have a smaller mass defect.
Justify your choice qualitatively, without using equations.

▶️Answer/Explanation

Ans:

(a) To find the mass defect, we first find the total constituent mass:
                                       88 protons: m = (88)(1.0078) = 88.6864 u
                                       138 neutrons: m = (138)(1.0087) = 139.2006 u
                                       M(total) = 227.8870 u

Then we have:
                                       Mass defect = 227.8870 u − 226.0244 u = 1.8626 u

(b) The mass defect represents the binding energy holding the nucleus together. This energy can be found by using Einstein’s equation \(E=mc^{2}\).

Using the conversion factor from the Table of Information for AP Physics 2:
                                       1 u = 931 MeV/\(c^{2}\) and \(E=mc^{2}\)

The total binding energy (BE) is
                                       BE = (931.5 MeV/u)(1.8626 u) = 1,735.0119 MeV

The binding energy per nucleon is
                                       BE/nucleons = 1,735.0119/226 = 7.677 MeV

(c)                                \(^{226}_{88}Ra\rightarrow ^{222}_{86}Rn+^{4}_{2}He+Q\)
    OR                          \(^{226}_{88}Ra\rightarrow ^{222}_{86}X+^{4}_{2}He+Q\)

(d) __X__ The reactants have a larger mass defect. Since this reaction is energy releasing, energy has left the nuclei. This energy has come from the mass, so the mass defect must be increased. The nucleons in the reactant side have greater binding energy on average than those on the product side.

Question

A student is investigating an unknown radioactive source with a Geiger counter. The Geiger counter clicks when it detects ionizing radiation but does not distinguish between alpha, beta, and gamma particles. The number of clicks indicates the number of ionizing events happening in the Geiger counter device due to the radiation it receives. The student puts a piece of paper between the radioactive source and the Geiger counter and does not notice a change in the clicking rate. However, the addition of one sheet of aluminum foil does decrease the rate. When several sheets of aluminum foil are layered in a thickness of about a centimeter, there are no clicks at all.
(a) What type of radiation is being emitted? Justify your choice by describing alpha, beta, and gamma particles and how they interact with matter.
(b) What is happening to the atomic number and mass number within each atom that emits this radiation? Is it possible to predict which radioactive atoms within the sample will be emitting the radiation next?
(c) Is the binding energy per nucleon within the radioactive atoms changing as the radiation is being emitted? If so, is the binding energy per nucleon going up or down? Justify your answer with a qualitative explanation.
(d) If the half-life of the radioactive material being investigated is 12 hours, what will happen to the click rate of the Geiger counter after 3 days?

▶️Answer/Explanation

Ans:

(a) Beta particles are being emitted. Alpha particles are stopped by paper, and gamma rays would penetrate the aluminum foil easily. Alpha particles are so easily stopped because they are big and highly charged. Beta particles have a single quantum of charge and are quite small, so they are harder to stop. Gamma rays are high frequency electromagnetic radiation and carry no charge. Thus they are the most difficult to stop.

(b) It is not possible to predict which specific radioactive atom will decay next. The process is random but defined by a certain probability. Upon emission of a beta particle, the atom will lose 4 units of mass and have its atomic number reduced by 2.

(c) The daughter nuclei will be more stable than before the beta particle was emitted. Therefore, its binding energy/nucleon will be higher. Another way to think about this is that the ejected electron is taking away energy in addition to its mass, leaving the daughter nuclei with a greater mass deficit per nucleon than before                                 \((E=mc^{2})\).

(d) With a half-life of 12 hours, 3 days equals 6 half-lives. Therefore, the amount of radioactive atoms in the sample will have been reduced by approximately \((1/2)^{6}\) after 3 days. As a result, the click rate on the Geiger counter would also be reduced by this factor (1/64).

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