Problem set 46

1. Parity non-conversion was established in $\beta$-decay when it was observed that from $Co^{60}$ nuclei:
1. Electrons were emitted equally in all directions
2. More electrons were emitted in direction opposite to that of magnetic field
3. Electrons were not emitted in any direction
4. More electrons were emitted perpendicular to the direction of magnetic field
Wu et. al. showed that parity is not conserved in weak interactions by observing the beta decay of $Co^{60}$. The spins of the $Co^{60}$ atoms were aligned in a particular direction, by aligning the electric dipole moments in a magnetic field at low temperature. The parity operation was accomplished by reversing the direction of the magnetic field. It was observed that the electrons were emitted in a preferred direction i. e. the direction opposite to the magnetic field, thereby violating parity. Thus it was concluded that parity is not conserved in weak interactions. It was also observed that the preferred direction for positrons is along the magnetic field. i. e. particles and anti-particle behave differently in beta decay, thereby violating the C symmetry. However, in ordinary beta decay, the combination CP is conserved.

Hence, answer is (B)

2. According to the liquid drop model, the occurrence of fission is due to competition between :
1. Surface energy term and symmetry energy term
2. Surface energy term and Coulomb energy term
3. Volume energy term and surface energy term
4. Volume energy term and Coulomb energy term
According to liquid drop model there is competition between the surface energy term and Coulomb energy term. The surface energy increases with deformation, and the Coulomb energy decreases.

Hence, answer is (B)

3. Binding energy difference in mirror nuclei can be understood using Coulomb energy difference. This indicates that :
1. Nuclear force is spin dependent
2. Nuclear force is strong force
3. Nuclear force is as strong as Coulomb force
4. Nuclear force is charge independent
“Mirror nuclei” are pairs of nuclei in which the proton number in one equals the neutron number in the other and vice versa. In one of the mirror nuclei the proton number is $Z=(A+1)/2$ and the neutron number is $N= (A−1)/2$; whilst in the other the proton number is $Z=(A−1)/2$ and the neutron number $N= (A+1)/2$. Examples are $^{13}_7N$ and $^{13}_6C$ or $^{31}_{16}S$ and $^{31}_{15}P$. The nuclei in these pairs differ from each other only in that a proton in one is exchanged for the neutron in the other. There is strong reason to believe that the force between nucleons does not differentiate between neutrons and protons; consequently, the “nuclear part” of the binding energy of two mirror nuclei must be the same. Hence, the mass difference of two mirror nuclei can be due only to the difference between the proton mass and the neutron mass, and the different Coulomb energies of the two.

This indicates that nuclear force is charge independent.

Hence, answer is (D)

4. Give the approximate values for the corresponding lifetimes of hadronic decay, electromagnetic decay and weak decay
1. $10^{-9}\:sec$, $10^{-6}\:sec$, $10^{-3}\:sec$
2. $10^{-12}\:sec$, $10^{-9}\:sec$, $10^{-6}\:sec$
3. $10^{-15}\:sec$, $10^{-13}\:sec$, $10^{-6}\:sec$
4. $10^{-23}\:sec$, $10^{-18}\:sec$, $10^{-10}\:sec$

(D) $10^{-23}\:sec$, $10^{-18}\:sec$, $10^{-10}\:sec$

5. Based on the additive quantum numbers such as Lepton number, Baryon number, charge of the particle and Isospin, indicate the following nuclear reaction cannot be induced with the following combination: $$n\rightarrow p+e^-+\nu_e^-$$
1. Q, B are conversed, but $I_3$, L are not conserved
2. Q, B, L are conversed, but $I_3$ is not conserved
3. Q, B, $I_3$ are conversed, but L is not conserved
4. B, $I_3$, L are conversed, but Q is not conserved

	$n\rightarrow p+e^-+\nu_e^-$
Q	$0\rightarrow 1-1+0$	Conserved
B	$1\rightarrow 1+0+0$	Conserved
L	$0\rightarrow 0+1-1$	Conserved
$I_3$	$-\frac{1}{2}\rightarrow \frac{1}{2}$	Not conserved

Hence, answer is (B)