Types of Radioactive Decay AP Physics 2 MCQ – Exam Style Questions etc.
Types of Radioactive Decay AP Physics 2 MCQ
Unit 15: Modern Physics
Weightage : 15–18%
Exam Style Practice Questions ,Types of Radioactive Decay AP Physics 2 MCQ
Question
Some scientists want to perform electron diffraction experiments on crystals that have an intermolecular spacing of about \(10^{-10}\) m. They have two processes by which they can produce energetic electrons that can strike the crystals.
Process 1 is the beta decay of potassium into calcium: 
The mass of 
is 39.964000 u and the mass of 
is 39.962591 u. As an approximation, assume that the mass of the neutrino is negligible and that all the decay energy is acquired by the electron.
Process 2 is the emission of electrons when electromagnetic radiation of wavelength 7.5 nm shines on a copper surface with work function 4.7 eV.
(a) Describe a single criterion for determining whether each of these processes can be used to produce electrons for the diffraction experiments.
(b) Determine whether each of these processes could be used to produce electrons appropriate for the diffraction experiments. Justify your answers mathematically.
Answer/Explanation
Ans:(a) For indicating an appropriate quantity to compare — either length or energy
For indicating the appropriate specic quantities to compare: de Broglie wavelengths of electrons and crystal spacing
OR
kinetic energy of electrons and kinetic energy needed to have a de Broglie wavelength equal to the crystal spacing
Example: The energy of electrons produced by each method is comparable to the energy of electrons with de Broglie wavelengths corresponding to the crystal spacing.
(b) For converting the electron energies to de Broglie wavelengths
OR
using the crystal spacing as a de Broglie wavelength to determine
the required electron energy Decay process
For determining the mass difference of the atoms
For converting the mass difference to energy
For correctly evaluating the energy against the crystal Photoelectric process
For determining the energy of the incident light
For determining the maximum kinetic energy K of the ejected electrons
For correctly evaluating the kinetic energy against the crystal
Example: Determine the electron kinetic energy needed for the electron to have a de Broglie
wavelength of\(10^{-10}\) m :
Determine the energy released by the decay reaction, which is converted to kinetic energy of the electron:
NOTE: is is four orders of magnitude larger than , so the wavelength will be two orders of magnitude smaller than desired. is process will not work well. Determine the maximum kinetic energy K of electrons emitted in the photoelectric effect: 
NOTE: is is comparable to 150 eV, so this process will work.
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.