nuclear

1. Which of the following matrices is an element of the group SU(2)
1. $\begin{pmatrix}1&1\\0&1\end{pmatrix}$
2. $\begin{pmatrix}\frac{1+i}{\sqrt{3}}&\frac{-1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}&\frac{1-i}{\sqrt{3}}\end{pmatrix}$
3. $\begin{pmatrix}2+i&i\\3&1+i\end{pmatrix}$
4. $\begin{pmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}$
The special unitary group of degree $n$, denoted $SU(n)$, is the Lie group of $n×n$ unitary matrices with determinant 1. For $2\times2$ matrix the group is denoted by $SU(2)$. Let $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. As $U$ is unitary, we have $UU^\dagger=U^\dagger U=1$. This gives $$aa^*+bb^*=1\quad(1)$$ $$cc^*+dd^*=1\quad(2)$$ $$ac^*+bd^*=0\quad(3)$$ $$ca^*+db^*=0\quad(4)$$ Also $$det~U=ad-bc=1\quad(5)$$ From (4), we have $$\frac{d}{c}=-\frac{a^*}{b^*}$$ substituting in (5) $$-(aa^*+bb^*)\frac{c}{b^*}=1$$ Using (1) we get $$c=-b^*$$ Similarly, we can show that $$d=a^*$$ Hence, in general the $2\times2$ matrix $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is of $SU(2)$ group if
• Its determinant is 1 and
• It is in the form $U=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix}$
These conditions are satisfied by $\begin{pmatrix}\frac{1+i}{\sqrt{3}}&\frac{-1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}&\frac{1-i}{\sqrt{3}}\end{pmatrix}$. Hence answer is (B).
2. The radius of $^{64}_{29}Cu$ nucleus is measured to be $4.8\times10^{-13} cm$.
1. The radius of $^{27}_{12}Mg$ nucleus can be estimated to be
1. $2.86\times10^{-13} cm$
2. $5.2\times10^{-13} cm$
3. $3.6\times10^{-13} cm$
4. $8.6\times10^{-13} cm$
2. The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as:
1. $A^{4/3}$
2. $A^{1/3}$
3. $A^{-1/3}$
4. $A^{-2/3}$
Nuclear radius is given by $R=R_0A^{1/3}$ $$R_{Cu}=R_0A_{Cu}^{1/3}\quad \text{and}\quad R_{Mg}=R_0A_{Mg}^{1/3}$$ $$\frac{R_{Mg}}{R_{Cu}}=\left(\frac{A_{Mg}}{A_{Cu}}\right)^{1/3}=\left(\frac{27}{64}\right)^{1/3}=\frac{3}{4}$$ \begin{align*} R_{Mg}&=\frac{3}{4}R_{Cu}\\ &=\frac{3}{4}\times4.8\times10^{-13}\\ &=3.6\times10^{-13} cm \end{align*} Hence, answer is (C)
The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as $A^{-1/3}$
Hence, answer is (C)
3. Deviation from Rutherford scattering formula for $\alpha$-particle scattering gives an estimate of:
1. Size of an atom
2. Thickness of target
3. Size of a nucleus
4. half life of $\alpha$-emitter
Rutherford formula is derived by assuming nucleus to be point charge. However, nucleus is not a point charge. Howver, nuclus is not a point charge. There is a charge distribution inside the nuclus. Due to which there is change in angular dependence of scattering. From this information one can estimate size of nucleus.
Hence, answer is (C)
4. The puzzle of magic numbers for nuclei was resolve by :
1. introducing hard-core potential
2. introducing Yukawa potential for shell model
3. introducing tensor character to nuclear force
4. introducing spin-orbit part in the nuclear potential
(D) introducing spin-orbit part in the nuclear potential
5. The ratio of the sizes (in terms of radii) of $^{208}_{82}Pb$ and $^{26}_{12}Mg$ nuclei is approximately:
1. 2
2. 4
3. $2\sqrt{2}$
4. 8
Nuclear radius is given by $R=R_0A^{1/3}$ $$R_{Pb}=R_0A_{Pb}^{1/3}\quad R_{Mg}=R_0A_{Mg}^{1/3}$$ \begin{align*} \frac{R_{Pb}}{R_{Mg}}&=\left(\frac{A_{Pb}}{A_{Mg}}\right)^{1/3}\\ &=\left(\frac{208}{26}\right)^{1/3}=2 \end{align*} Hence, answer is (A)
6. What is the energy of a gamma radiation backscattered at an angle $180^o$, if the incident energy is 10 MeV?
1. 10 MeV
2. 5 MeV
3. 0.511 MeV
4. 0.25 MeV
If a photon with energy $E_0$ strikes a stationary electron, then the energy of the scattered photon, $E$, depends on the scattering angle, $\theta$, that it makes with the direction of the incident photon according to the following equation: $$\left[\frac{1}{E}-\frac{1}{E_0}\right]=\frac{1}{m_ec^2}(1-\cos\theta)$$ $m_e c^2 = 511 keV = 0.511 MeV$ \begin{align*}\left[\frac{1}{E}-\frac{1}{10}\right]&=\frac{1}{0.511}(1-\cos180)\\ &=3.913894325 \end{align*} \begin{align*} \frac{1}{E}&=3.913894325-0.1\\ &=4.02213894325 \end{align*} $$E=0.25 MeV$$ Hence, answer is (D)
7. Ionisation chamber is effectively used for the measurement of :
1. Radiation
2. Radiation Dose
3. strength of radiation
4. Energy of radiation
Ionisation chamber is effectively used for the measurement of Radiation Dose as they provide an output that is proportional to dose.
Hence, answer is (B)
8. A satisfactory quenching gas in G.M. tube must have the following property:
1. Ionisation potential should be equal to the main counting gas in the tube
2. Ionisation potential should be higher than that of main counting gas in the tube
3. It must have very narrow ultraviolet absorption bands
4. When in excited state it must prefer to dissociate rather than to de-excite by the emission of photon
Quenching gas must have following properties:
1: Its ionisation potential should be less than the main gas
2: It should be complex and prone to dissociation rather than de-excite by photon emission
3: It should have sharp and intense absorption band in UV so that it quickly dissociates by absorbing undesired UV causing spurious discharge.

9. 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)

10. 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)

11. 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)

12. 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) $10^{-23}\:sec$, $10^{-18}\:sec$, $10^{-10}\:sec$

13. 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)

14. State the following decay mode in the category of allowed, forbidden and Fermi or Gammow-Teller transition
    
1. Fermi transition and allowed
2. Fermi transition and second forbidden
3. G-T transition and first forbidden
4. G-T transition and allowed
Selection rules for allowed transitions in $\beta$-decay are: \begin{equation*} \left.\begin{aligned} &\Delta I=0\\ &{\scriptstyle I_i=0\rightarrow I_f=0\quad \text{allowed}}\\ &\Delta\pi=0 \end{aligned}\right\}\text{Fermi rule } \end{equation*} \begin{equation*} \left.\begin{aligned} & \Delta I=0,\pm1\\ & {\scriptstyle I_i=0\rightarrow I_f=0\quad \text{forbidden}}\\ &\Delta\pi=0 \end{aligned}\right\}\text{G.T. rule} \end{equation*} where $I$ is nuclear spin and $\pi$ is parity.

Using these selection rule we can see that the transition is Fermi transition and allowed

Hence, answer is (A)

15. State the following decay mode in the category of allowed, forbidden and Fermi or Gammow-Teller transition
    
1. Fermi transition and allowed
2. Fermi transition and second forbidden
3. G-T transition and allowed
4. G-T transition and third forbidden
Using transitions we can see that the transition is G-T transition and allowed

Hence, answer is (C)

16. Consider a energy level diagram shown below, which corresponds to the molecular nitrogen
    
    If the pump rate $R$ is $10^{20}$ atoms cm^-3s^-1 and the decay routes are as shown with $\tau_{12}=20\:ns$ and $\tau_1=1\:\mu s$, the equilibrium populations of states 2 and 1 are, respectively,
1. $10^{14}\: cm^{-3}$ and $2\times10^{12}\: cm^{-3}$
2. $2\times10^{12}\: cm^{-3}$ and $10^{14}\: cm^{-3}$
3. $2\times10^{12}\: cm^{-3}$ and $2\times10^{6}\: cm^{-3}$
4. zero, and $10^{20}\: cm^{-3}$
The differential rate equations can be written as $$\frac{dN_2}{dt}=R-\frac{N_2}{\tau_{12}}$$ $$\frac{dN_1}{dt}=\frac{N_2}{\tau_{12}}-\frac{N_1}{\tau_{1}}$$ For equlibrium, we have $$\frac{dN_2}{dt}=\frac{dN_1}{dt}=0$$ $$R-\frac{N_2}{\tau_{12}}=0$$ $$\frac{N_2}{\tau_{12}}-\frac{N_1}{\tau_{1}}=0$$ \begin{align*} N_2&=R\tau_{12}\\ &=20\times 10^{-9}\times 10^{20}\\ &=2\times10^{12}\:cm^{-3} \end{align*} \begin{align*} N_1&=\frac{\tau_{1}N_2}{\tau_{12}}\\ &=\frac{1\times10^{-6}\times2\times10^{12}}{20\times10^{-9}}\\ &=10^{14}\:cm^{-3} \end{align*}

Hence, answer is (B)

17. Consider following particles: the proton $p$, the neutron $n$, the neutral pion $\pi^0$ and the delta resonance $\Delta^+$. When ordered of decreasing lifetime, the correct arrangement is as follows
1. $\pi^0, n, p, \Delta^+$
2. $ p,n, \Delta^+, \pi^0$
3. $ p,n,\pi^0, \Delta^+ $
4. $\Delta^+,n, \pi^0, p $
Mean-life: Proton: stable, Neutron: $885.7\pm0.8$ seconds, $\pi^0$ $0.84\times10^{-16}, $\Delta^+$: $6\times10^{-24}$

Answer is (C)

18. Weak nuclear forces act on
1. both hadrons and leptons
2. hadrons only
3. all particles
4. all charged particles

Weak nuclear forces act on both hadrons and leptons, while strong nuclear force acts on hadrons only.

Hence, answer is (A)

19. Which one of the following disintegration series of the heavy elements will give $^{209}Bi$ as a stable nucleus?
1. Thorium series
2. Neptunium series
3. Uranium series
4. Actinium series

Uranium series terminates with lead-206

Actinium series terminates with lead-207

Thorium series terminates with lead-208

Neptunium series terminates with bismuth-209 and thallium-205

Hence, answer is (B)

20. The order of magnitude of the binding energy per nucleon in a nucleus is
1. $10^{-5}$ MeV
2. $10^{-3}$ MeV
3. 0.1 MeV
4. 10 MeV

Hence, answer is (D)

21. The interaction potential between two quarks, separated by a distance $r$ inside a nucleon, can be described by ($a$, $b$ and $\beta$ are positive constants)
1. $ae^{-\beta r}$
2. $\frac{a}{r}+br$
3. $-\frac{a}{r}+br$
4. $\frac{a}{r}$

The interaction potential between two quarks, separated by a distance $r$ inside a nucleon, can be described by $\frac{a}{r}+br$

Hence, answer is (B)

22. Of the nuclei of mass number $A=125$, the binding energy calculated from the liquid drop model (given that the coefficients for the Coulomb and the asymmetry energy are $a_c=0.7$ MeV and $a_{sym}=22.5$ MeV respectively) is a maximum for
1. $^{125}_{54}$Xe
2. $^{125}_{53}$I
3. $^{125}_{52}$Te
4. $^{125}_{51}$Sb

Using relation $$Z=\frac{4a_{sym}+a_cA^{-1/3}}{8a_{sym}A^{-1}+2a_cA^{-1/3}}$$ we get $Z=52$

Hence, answer is (C)

23. Consider the following processes involving free particles
    1. $\bar n\rightarrow \bar p+e^++\bar \nu_e$
    2. $\bar p+n\rightarrow \pi^-$
    3. $ p+n\rightarrow \pi^++\pi^0+\pi^0$
    4. $p+\bar \nu_e\rightarrow n+e^+$
    Which of the following statements is true?
1. Process (i) obeys all conservation laws
2. Process (ii) conserves baryon number, but violates energy-momentum conservation
3. Process (iii) is not allowed by strong interactions, but is allowed by weak interactions
4. Process (iv) conserves baryon number, but violates lepton number conservation

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

Similarly, one can check for other process. Process (ii) conserves baryon number, but violates energy-momentum conservation

Hence, answer is (B)

24. Consider the energy level diagram shown below, which corresponds to the molecular nitrogen laser.
    
    If the pump rate $R$ is $10^{20}$ atoms cm$^{-3}$ s$^{-1}$ and the decay routes are as shown with $\tau_{21} =20\: ns$ and. $\tau_{1} = 1\: \mu s$, the equilibrium populations of states 2 and 1 are, respectively,
1. $10^{14}$ cm$^{-3}$ and $2\times10^{12}$ cm$^{-3}$
2. $2\times10^{12}$ cm$^{-3}$ and $10^{14}$ cm$^{-3}$
3. $2\times10^{12}$ cm$^{-3}$ and $2\times10^{6}$ cm$^{-3}$
4. zero and $10^{20}$ cm$^{-3}$

$$\frac{dN_2}{dt}=R-\frac{N_2}{\tau_{21}}$$ and $$\frac{dN_1}{dt}=\frac{N_2}{\tau_{21}}-\frac{N_1}{\tau_{1}}$$ At equilibrium $\frac{dN_2}{dt}=\frac{dN_1}{dt}=0$ $$N_2=\tau_{21}R=2\times10^{12}\: cm^{-3}$$ $$N_1=\frac{\tau_1N_2}{\tau_{21}}=10^{14}\:cm^{-3}$$

Hence, answer is (B)

25. What are the expected types of gamma ray transitions between the following states of odd ‘A’ nuclei : $$g_{9/2}\rightarrow P_{1/2}$$
1. M4 and E5
2. M1 and E2
3. M3 and E4
4. M6 and E7

Gamma rays emitted in electric $E$-$l$ transitions carry off angular momentum $l$ and parity $(-1)^l$. Therefore, angular momenta and parity of final and initial states are related by $$\Delta I=|I_i-I_f| \text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^l$$ The selection rules for magnetic $M$-$l$ transitions are $$\Delta I=|I_i-I_f|\text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^{l-1}$$ Following table shows selections rules for $\gamma$-ray transitions

Type	$l=\Delta I$	$\Delta\pi$
E1	1	yes
M1	1	no
E2	2	no
M2	2	yes
E3	3	yes
M3	3	no
E4	4	no
M4	4	yes
E5	5	yes
M5	5	no

For $g_{9/2}\rightarrow P_{1/2}$ $$\Delta I=l=\left|\frac{9}{2}-\frac{1}{2}\right|\text{ to } \frac{9}{2}+\frac{1}{2}=4, 5$$ Also, we know that the state with even $l$ value have even parity and the state with odd $l$ have odd parity. Hence, for $g_{9/2}$ parity is even and for $P_{1/2}$ parity is odd. Hence, there is parity change i.e. $$\Delta\pi=\text{ yes}$$ For $l=4,5$ and $\Delta\pi=\text{ yes}$, from above table we get that transitions as M4 and E5

Hence, answer is (A)

26. What are the expected types of gamma ray transitions between the following states of odd ‘A’ nuclei : $$f_{5/2}\rightarrow P_{3/2}$$
1. E3, M4, E5, M6
2. M1, E2, M3, E4
3. M4, E5, M6, E7
4. M4, M1, E3, E4

Gamma rays emitted in electric $E$-$l$ transitions carry off angular momentum $l$ and parity $(-1)^l$. Therefore, angular momenta and parity of final and initial states are related by $$\Delta I=|I_i-I_f| \text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^l$$ The selection rules for magnetic $M$-$l$ transitions are $$\Delta I=|I_i-I_f|\text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^{l-1}$$ Following table shows selections rules for $\gamma$-ray transitions

Type	$l=\Delta I$	$\Delta\pi$
E1	1	yes
M1	1	no
E2	2	no
M2	2	yes
E3	3	yes
M3	3	no
E4	4	no
M4	4	yes
E5	5	yes
M5	5	no

For $f_{5/2}\rightarrow P_{3/2}$ $$\Delta I=l=\left|\frac{5}{2}-\frac{3}{2}\right|\text{ to } \frac{5}{2}+\frac{3}{2}$$ $$\Delta I=l=1,2,3,4$$ Also, we know that the state with even $l$ value have even parity and the state with odd $l$ have odd parity. Hence, for $f_{5/2}$ parity is odd and for $P_{3/2}$ parity is odd. Hence, there is no parity change i.e. $$\Delta\pi=\text{ no}$$ For $l=1,2,3,4$ and $\Delta\pi=\text{ no}$, from above table we get that transitions as M1, E2, M3 and E4

Hence, answer is (B)

27. Based on the additive quantum numbers such as strangeness, Baryon number, charge of the particle and Isospin, indicate the following nuclear reaction cannot be induced with the following combination: $$\pi^++n\rightarrow \pi^0+k^+$$
1. Q, B, S are conversed, but $I_3$ is not conserved
2. Q, B are conversed, but S, $I_3$ are not conserved
3. Q, $I_3$ are conversed, but B, S are not conserved
4. B, S, $I_3$ are conversed, but Q is not conserved

	$\pi^++n\rightarrow \pi^0+k^+$
Q	$1+0\rightarrow 0+1$	Conserved
B	$0+1\rightarrow 0+0 $	Not Conserved
S	$0+0\rightarrow 0+1$	Not Conserved
$I_3$	$1+\frac{1}{2}\rightarrow 1+\frac{1}{2}$	Conserved

Hence, answer is (C)

28. Magnetic moment of duteron $\mu_D\ne \mu_p+\mu_n$. This is due to:
1. Spin dependence of nuclear force
2. Tensor character of nuclear force
3. Spin-orbit force part of nuclear force
4. Hard core part of the nuclear force

The deuteron consists of a neutron and a proton. The deuteron is known to have a quadrupole moment, 0.00286 barns, which tells us that the deuteron is not perfectly spherical and that the force between two nucleons is not spherically symmetric. The force between two nucleons has two components, a spherically symmetric central force and an asymmetric tensor force that depends on the angles between the spin axis of each nucleon and the line connecting.

Hence, answer is (B)

29. It is required to operate a G.M. counter with a maximum radial field $10^7$ V/m. The applied voltage required if the radii of the wire and tube are 0.002 cm and 1 cm respectively.
1. $10^7$ volts
2. $1242\times 10^7$ volts
3. $1242$ volts
4. $12$ volts

In a cylindrical geometry with the anode at the center, the electric field at radius $r$ from the anode is given by $$E(r)=\frac{V}{r\ln{\left(\frac{b}{a}\right)}}$$ where $V =$ the applied voltage, $a =$ anode wire radius, $b =$ cathode tube inner radius. $$V=E(r)\times r\times\ln{\left(\frac{b}{a}\right)}$$ $E(r)$ is maximum on the anode surface, hence, $r=0.002 \:cm=2\times10^{-5}m$ $$V=10^{7}\times 2\times10^{-5}\times\ln{\left(\frac{1}{0.002}\right)}$$ $$V=1242\:volts$$

Hence, answer is (C)

30. NaI(Tl) scintillation detector is used only for the detection of gamma radiation because
1. it has large scattering cross-section
2. it has small Comption scattering
3. it has small absorption cross-section
4. it has large absorption cross-section

It has large absorption cross-section

Hence, answer is (D)

31. In nuclear direct reactions, time of interaction is of the order of :
1. $10^{-10}$ sec
2. $10^{-16}$ sec
3. $10^{-22}$ sec
4. $10^{-30}$ sec

In nuclear direct interaction time must be very short :$10^{-22}$ sec

In compound nuclear interaction time is large :$10^{-18}$ to $10^{-16}$ sec

Hence, answer is (C)

32. The following decay states a conservation law that forbids it because: $$n\rightarrow p+e^-$$
1. conservation of angular momentum and conservation of Lepton numbers are both violated
2. conservation of baryon number and conservation of Lepton number are both violated
3. conservation of energy is violated
4. conservation of electric charge is violated

$$n\rightarrow p+e^-$$

1. There is conversation of electric charge as protron has charge $+e$, electron has charge $-e$ and neutron is charge-less.
2. baryon number is conserved since baryon number of proton is $+1$, of neutron is $+1$ and of electron is $0$
3. conservation of Lepton number is violated since Lepton number of proton is $0$, of neutron is $0$ and of electron is $1$
4. conservation of angular momentum is also violated since, when the decay of the neutron into a proton and an electron was observed, it did not fit the pattern of two-particle decay. That is, the electron emitted does not have a definite energy as is required by conservation of energy and momentum for a two-body decay.

Hence, answer is (A)

33. Let us approximate the nuclear potential in the shell model by a three dimensional isotropic harmonic oscillator. Since the lowest two energy levels have angular momenta $l=0$ and $l=1$ respectively, which of the following two nuclei have magic numbers of protons and neutrons?
1. $_2^4He$ and $_8^{16}O$
2. $_1^2D$ and $_4^{8}Be$
3. $_2^4He$ and $_4^{8}Be$
4. $_2^4He$ and $_6^{12}C$

The energy eigenvalues of the harmonic oscillator are $$E_n=(n+\frac{1}{2})\hbar\omega\quad n=1,2,3,\cdots$$ In case of harmonic oscillator for a given $n$ $$l=n-1,n-2,\cdots0$$ $$m_l=-l,-l+1,\dots l$$ $$m_s=\pm\frac{1}{2}$$ For $l=0$, $n=1$, $m_l=0$ and $m_s=\pm\frac{1}{2}$. Hence there are two states explaining the first magic number, 2. Two nucleons of a given type can occupy the lowest energy level. $_2^4He$ has the lowest energy level for protons completely filled with its two protons, and the lowest level for neutrons completely filled with its two neutrons. That makes $_2^4He$ the first doubly-magic nucleus.

For $l=1$, $n=2$, $m_l=-1, 0,1$ and $m_s=\pm\frac{1}{2}$. Therefore, $l=1$ at $n=2$ corresponds to 6 energy states. Combined with the two $l= 0$ states at energy level $n=1$, that gives a total of 8. The second magic number 8 has been explained! It requires 8 nucleons of a given type to fill the lowest two energy levels. It makes $_8^{16}O$ with 8 protons and 8 neutrons the second doubly-magic nucleus.

Hence, answer is (A)

34. The reaction $^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$ cannot proceed via strong interactions because it violets the conservation of
1. angular momentum
2. electric charge
3. baryon number
4. isospin

A nucleus of deuteron contains one proton and one neutron (each with a baryon number of 1) and has a baryon number of 2 and charge +1. isospin=0, Owing to the generalized Pauli principle, it must have a symmetric spin state, i.e. $S= 1$, and a total angular momentum $J=1$. Helium nucleus is composed of two protons and two neutrons. Hence, Baryon number 4 and charge +2, isospin=0

	$^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$
Q	$1+1\rightarrow 2+0$	Conserved
B	$2+2\rightarrow 4+0$	Conserved
Isospin	$0+0\rightarrow 0+1$	Not Conserved

Hence, answer is (D)

35. The experimentally measured spin $g$ factors of a proton and a neutron indicate that
1. both proton and neutron are elementary point particles
2. both proton and neutron are not elementary point particles
3. while proton is an elementary point particle, neutron is not
4. while neutron is an elementary point pat1icle, proton is not

both proton and neutron are not elementary point particles

Hence, answer is (B)

36. Let $E_s$ denote the contribution of the surface energy per nucleon in the liquid drop model. The ratio $E_s\left(^{27}_{13}Al\right):E_s\left(^{64}_{30}Al\right)$ is
1. 2:3
2. 4:3
3. 5:3
4. 3:2

Surface energy $$E_s\propto A^{2/3}$$ Surface energy per nucleon is $$E_s\propto \frac{A^{2/3}}{A}=\frac{1}{A^{1/3}}$$ $$\frac{E_s\left(^{27}_{13}Al\right)}{E_s\left(^{64}_{30}Al\right)}=\frac{64^{1/3}}{27^{1/3}}=\frac{4}{3}$$

Hence, answer is (B)

37. According to the shell model, the nuclear magnetic moment of the $^{27}_{13}Al$ nucleus is (Given that for a proton $g_l=1$, $g_s=5.586$, and for a neutron $g_l=0$, $g_s=-3.826$)
1. $-1.913\mu_N$
2. $14.414\mu_N$
3. $4.793\mu_N$
4. 0

In $^{27}_{13}Al$ there are odd numbers of protons (13) and even numbers of neutrons (14). The unpaired proton is responsible for magnetic moment. $^{27}_{13}Al$ has an odd proton in the $j = 5/2 1d (l=2)$ state. Magnetic moment for single nucleon is given by $$\mu_j=\left(g_ll+\frac{1}{2}g_s\right)\mu_N$$ $$\mu_j=\left(1\times2+\frac{1}{2}5.586\right)\mu_N$$ $$\mu_j=4.793\mu_N$$

Hence, answer is (C)