Unit 15: Modern Physics
Weightage : 15–18%
\(^{214}_{83}\mathrm{Bi}\rightarrow X+\beta^{+}\)
In the decay represented above, what is the nucleon (mass) number of particle \(X\)?
Correct Answer: \( \boxed{\mathrm{D}} \)
In positron (\(\beta^{+}\)) decay, a proton is converted into a neutron:
\(p\rightarrow n+\beta^{+}+\nu_e\)
During this process:
• The mass number (nucleon number) remains unchanged.
• The atomic number decreases by 1.
Therefore,
\(^{214}_{83}\mathrm{Bi}\rightarrow{}^{214}_{82}\mathrm{Pb}+\beta^{+}\)
Thus, particle \(X\) has a mass number of
\(A=214\)
Therefore, the correct answer is (D).
A scientist is studying the radioactive decay of an element and wants to determine the sign of some of the particles emitted. The scientist lets a beam of the particles, all with the same speed, pass through an opening in a shield and enter the field created by a wire carrying a constant current. Paths 1 and 2 represent the observed paths of the particles. Based on the paths, which of the following could be the types of particles?
Path 1 Particle Path 2 Particle
(A) Electron Alpha
(B) Alpha Electron
(C) Positron Alpha
(D) Alpha Positron
Correct Answer: \( \boxed{\mathrm{C}} \)
The magnetic force on a moving charged particle is given by
\( \vec{F}=q\vec{v}\times\vec{B} \)
Using the right-hand rule, both particles curve in the same direction, so they must have the same sign of charge. Since both an alpha particle and a positron are positively charged, they are the only possible pair.
The radius of curvature is determined by
\( r=\dfrac{mv}{qB} \)
The alpha particle has approximately four times the mass and twice the charge of a positron. Therefore,
\( r_{\alpha}\approx\dfrac{4m}{2q}\cdot\dfrac{v}{B}=2r_{\mathrm{positron}} \)
Thus, the alpha particle follows the path with the larger radius of curvature (Path 2), while the positron follows the tighter curve (Path 1).
Therefore, the correct answer is (C).
In the reaction below, what is the mass number for particle \(X\)?
\(^{27}_{13}\mathrm{Al}+{}^{4}_{2}\mathrm{He}\rightarrow{}^{30}_{15}\mathrm{P}+X\)
Correct Answer: \( \boxed{\mathrm{A}} \)
In a nuclear reaction, the total mass number must be conserved.
Initial mass number:
\(27+4=31\)
Final mass number:
\(30+A_X=31\)
Therefore,
\(A_X=31-30=1\)
Thus, the mass number of particle \(X\) is 1.
Therefore, the correct answer is (A).
