Fission, Fusion, and Nuclear Decay AP Physics 2 MCQ – Exam Style Questions etc.
Fission, Fusion, and Nuclear Decay AP Physics 2 MCQ
Unit 15: Modern Physics
Weightage : 15–18%
Exam Style Practice Questions ,Fission, Fusion, and Nuclear Decay AP Physics 2 MCQ
Question
A sample of radioactive sources is enclosed within a lead-shielded container with a narrow exit aperture that ensures that any ejected particles will exit the container directly to the right. The ejected particles pass between charged parallel conductive plates and a region of magnetic field that is directed into the page. Three particles exit the container and follow paths as shown in the figure. Students observing the particles make these statements:
Student A: Both particles 2 and 3 pass through the capacitor region undeflected so there is no net force on them. They both must be neutral particles.
Student B: The path of particle 3 implies it has a negative charge. Therefore, the bottom plate of the capacitor must be negatively charged.
Student C: Particle 1 is positive because it curves upward.
(A) List all parts of the students’ statements that are correct. Explain your reasoning for each.
(B) List all parts of the students’ statements that are incorrect. Explain your reasoning for each.
( C) On the figure, sketch the electric field vectors between the capacitor plates that are consistent with the motion of the particles.
(D) Particle 3 is detected at distance D from its exit point from the capacitor plates. Using this information, derive an expression for the charge to mass ratio (q/m) of particle 3 in terms of D, E (the electric field between the capacitor plates), and B (the magnetic field).
(E) Using the equation derived in (D), a charge to mass ratio (q/m) of 1.76 x \(10^{11}\) C/kg was calculated. Explain how this number could be used to determine the set of possible masses of particle 3.
(F) One of the radioactive sources enclosed in the shielded container is fermium-257, which has 100 protons and a half-life of 100.5 days. Fermium transmutes into Californium (Cf) by emitting an alpha particle with a velocity of 2.0 x \(10^{7}\) m/s.
(i) Write the complete nuclear equation of this decay reaction.
(ii) Write, but do not solve, a symbolic expression that could be used to calculate the mass released to energy in one single fermium-257 decay.
(iii) Assuming the fermium-257 is isolated and stationary, calculate the velocity of the Californium nucleus after the alpha particle is ejected.
Answer/Explanation
Ans:
Part (A)
Student A is correct that particles 2 and 3 have no net force acting on them while between the charged plates. This is evident in the fact that they travel in a straight line through the charged plate.
AND
Student A is correct in stating that particle 2 is neutral as it travels in a straight line through the magnetic field. Knowing that a charged particle will experience a force while passing through a magnetic field, particle 2 must be uncharged.
Student B is correct in stating that particle 3 must have a charge, since a moving charged particle will experience a force when passing through a magnetic field. Student B is also correct in stating that the path implies a negative charge. According to the right-hand rule for a charged particle in a magnetic field, the direction of force on a moving charged particle will be perpendicular to both the magnetic field and the direction of positive charge motion. Using this right-hand rule, we find a circular path directed upward for a positive charge, thus the downward curve of particle 3 suggests a negative charge.
AND
Student B is correct in stating that the bottom plate must be negatively charged. Since particle 3 is negative as previously stated, the negative charge will experience a downward force from the magnetic field. In order to maintain a straight line and thus no acceleration, particle 3 must experience an upward force to balance the downward force. This can be achieved only if the bottom plate repels the negative charge upward, indicating that the bottom plate is negatively charged.
Part (B)
Student A is incorrect in stating that particle 3 is neutral as evidenced by the fact that it experiences a force exerted by the magnetic field and arcs downward as soon as it leaves the electric field between the charged plates.
Student C is incorrect in stating that particle 1 is positively charged. Although the direction of the magnetic force on a positive charge would be upward, the bottom plate is negative and the top plate is positive. Thus, the electric force on particle 1 is downward. It is not possible to tell from the given information what the charge of particle 1 is. We can only say for sure that particle 1 has a net charge.
Part (C)
Part (D)
Substituting in the equation for velocity from above: 
Part (E)
Example: The mass can be found by dividing the charge by the charge to mass ratio (q/m). However, we know that particles come only in multiples of the electron charge. So, dividing the multiple of the electron charge by the charge to mass ratio will give us a set of possible masses for particle 3.
Part (F)
Question
An unstable nucleus that is initially at rest decays into a nucleus of fermium-252 containing 100 protons and 152 neutrons and an alpha particle that has a kinetic energy of 8.42 MeV. The atomic masses of helium-4 and fermium-252 are 4.00260 u and 252.08249 u, respectively.
a. What is the atomic number of the original unstable nucleus?
b. What is the velocity of the alpha particle?
c. Where does the kinetic energy of the alpha particle come from? Explain briefly.
d. Assuming all of the energy released in the reaction is in the form of kinetic energy of the alpha particle, determine the exact mass of the original unstable nucleus, to an accuracy of 3 thousandths of a decimal.
e. Suppose that the fermium-252 nucleus could undergo a decay in which a β– particle was produced. How would this affect the atomic number of the nucleus? Explain briefly.
Answer/Explanation
Ans:
a) The reaction can be written as follow: ? → 252Fm100 + 4He2. For nucleons to add up properly, the original nucleus must have been 256X102 (this is called Nobelium. FYI)
b) K = ½ mv2 … 8.42×106 eV *1.6×10–19 J/eV = ½ (4.0026u*1.66×10–27 kg) v2 … v = 2.014×107 m/s
c) The kinetic energy comes from the conversion of mass to energy in the reaction. The mass before the reaction and the mass after the reaction are unequal, this is known as the mass difference. The energy equivalent of this mass difference contributes to the kinetic energy of the alpha particle.
d) Converting the alpha particle energy into mass equivalence. 8.42 Mev / (931 MeV /u) = 0.009 u. Adding the masses of all the products. 252.08249 + 4.0026 + 0.009 u = 256.094 u
e) In B– decay, a neutron turns into a proton and releases an electron beta particle. Since there is now one more proton, the atomic number increases by 1.