Problem set 48

1. 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)

2. 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)

3. The short wavelength cut-off of the continuous X-ray spectrum from a nickel target is 0.0825 nm. The voltage required to be applied to the X-ray tube must be:
1. 0.15 kV
2. 1.5 kV
3. 15 kV
4. 150 kV
According to Duane–Hunt law $$\lambda_{min}=\frac{1239.8}{V \text{ in }kV}pm$$ $$V \text{ in }kV=\frac{1239.8}{\lambda_{min}\text{ in }pm}$$ $$V \text{ in }kV=\frac{1239.8}{82.5}=15\:kV$$ Hence, answer is (C)

4. If the Planck's constant were to be zero, then the total energy contained in a box filled with radiation of all frequencies at temperature T would be (where $k$ is Boltzmann constant and $T\ne0$)
1. zero
2. infinite
3. $\frac{3}{2}kT$
4. $kT$
According to Stefan-Boltzmann law the the energy flux $\Phi$ emitted from a blackbody at temperature $T$, is $\Phi = \sigma T^4$, where $\sigma=\frac{2\pi^5k^4}{15h^3c^2}$. AS $\sigma\propto\frac{1}{h^3}$, energy will be infinite when $h=0$

Hence, answer is (B)

5. The number of electrons per unit volume in metallic potassium is $1.33\times 10^{28}\: atoms/m^3$ and if each potassium atom donates one electron to the electron gas, its Fermi energy in eV is :
1. 4.72
2. 3.23
3. 2.05
4. 1.85
$$E_F=\frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{\frac{2}{3}}$$ \begin{align*}&E_F=\frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{\frac{2}{3}}\\ &={\textstyle \frac{\left(6.62607004 \times 10^{ −34}\right)^2}{8\times 9.109\times10^{−31}}\left(\frac{3\times1.33\times 10^{28}}{3.14}\right)^{\frac{2}{3}}J}\\ &=3.28446\times10^{-19}J\\ &=\frac{3.28446\times10^{-19}}{1.6\times10^{-19}}\\ &=2.05 eV \end{align*} Hence, answer is (C)

Problem set 47

1. The Lagrangian of a particle moving in a plane is given in Cartesian coordinates as $$L=\dot x\dot y-x^2-y^2$$ In polar coordinates the expression for the canonical momentum (conjugate to the radial coordinate ) is
1. $\dot r\sin\theta+r\dot\theta\cos\theta$
2. $\dot r\cos\theta+r\dot\theta\sin\theta$
3. $2\dot r\cos{2\theta}-r\dot\theta\sin{2\theta}$
4. $\dot r\sin{2\theta}+r\dot\theta\cos{2\theta}$

Substituting $x=r\cos\theta$ and $y=r\sin\theta$ in $L$ \begin{align*} L&={\scriptstyle\left(\dot r\cos\theta-r\dot\theta\sin\theta\right)\left(\dot r\sin\theta+r\dot\theta\cos\theta\right)}\\ &-r^2\cos^2\theta-r^2\sin^2\theta\\ &=\dot r^2\cos\theta\sin\theta+r\dot r\dot\theta\cos^2\theta\\ &-r\dot r\dot\theta\sin^2\theta- r^2\dot\theta^2\cos\theta\sin\theta-r^2\\ &=\frac{1}{2}\dot r^2\sin{2\theta}+r\dot r\dot\theta\cos{2\theta}\\ &-\frac{1}{2} r^2\dot\theta^2\sin{2\theta}-r^2 \end{align*} \begin{align*} p_r=\frac{\partial L}{\partial\dot r}=\dot r\sin{2\theta}+r\dot\theta\cos{2\theta} \end{align*}

Hence, answer is (D)

2. The Hermite polynomial $H_n(x)$ satisfies the differential equation $$\frac{d^2H_n}{dx^2}-2x\frac{dH_n}{dx}+2nH_n(x)=0$$ . The corresponding generating function $G(x,t)=\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n(x)t^n$ satisfies the
1. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}=0$
2. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}-2t^2\frac{\partial G}{\partial t}=0$
3. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial G}{\partial t}=0$
4. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial^2 G}{\partial x\partial t}=0$
Let us consider option (A)
$\frac{\partial^2G}{\partial x^2}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{d^2H_n}{d x^2}t^n$, $\frac{\partial G}{\partial x}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{dH_n}{d x}t^n$, $2t\frac{\partial G}{\partial t}=2n\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n( x)t^n$ \begin{align*} &\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}\\ &{\scriptstyle =\sum\limits_{n=0}^{\infty}\frac{1}{n!}\left(\frac{d^2H_n}{d x^2}-2x\frac{1}{n!}\frac{dH_n}{d x}+2n\frac{1}{n!}H_n( x)\right)t^n}\\ &=0 \end{align*} Hence, answer is (A)

3. A sinusoidal signal of peak to peak amplitude 1V and unknown time period is input to the following circuit for 5 seconds duration. If the counter measures a value (3E8)_H in hexadecimal then the time period of the input signal is
   
1. 2.5 ms
2. 4 ms
3. 10 ms
4. 5 ms

In general n-bit counter counts $2^n$ numbers in a sequence ranging from 0 to $2^n-1$ and the counter cycles through the same sequence of numbers continuously so long as there is an incoming clock pulse. For one complete cycle we require $2^n-1$ pulses.

If $T$ is time period of a pulse then time required for one complete cycle of counts = $T\times(2^n-1)$.

Now, to count a decimal number $N$ we require $N$ pulses. This can be seen from a table below for a 2-bit counter.

Input pulse	Count	decimal equivalent
0	00	0
1	01	1
2	10	2
3	11	3

The decimal equivalent of $(3E8)_H={\scriptstyle 3\times16^2+14\times16^1+8\times16^0}=1000$ Hence, 1000 pulses of time period $T$ will be required to count the number $(3E8)_H$ in 5 seconds. $$5 \:sec=1000\times T$$ $$T=\frac{5}{1000}=5\:msec$$

Hence, answer is (D)

4. For a dynamical system governed by the equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, with $|x|\leq1$,
1. $x=-1$ and $x=1$ are both unstable fixed points
2. $x=-1$ and $x=1$ are both stable fixed points
3. $x=-1$ is an unstable fixed point and $x=1$ is a stable fixed point
4. $x=-1$ is a stable fixed point and $x=1$ is an unstable fixed point
Let $\frac{dx}{dt}=2\sqrt{1-x^2}=f(x)$

As $f(-1)=0$, x=-1 is a fixed point.

As $f(1)=0$, x=1 is a fixed point. $$f'(x)=\frac{-2x}{\sqrt{1-x^2}}$$

As $f'(-1)=\frac{2}{0}=\infty > 0$, $x=-1$ is a unstable point.

As $f'(1)=\frac{-2}{0}=-\infty < 0$, $x=1$ is a stable point.

Hence, answer is (C)

5. The value of the integral $\int_0^8\frac{1}{x^2+5}dx$, evaluated using Simpson’s $\frac{1}{3}$ rule with $h=2$, is
1. 0.565
2. 0.620
3. 0.698
4. 0.736
According to Simpson's 1/3 rule \begin{align*} I&=\frac{h}{3}\left[y_0+y_n\right.\\ &+4\left(y_1+y_3+\dots+y_{n-1}\right)\\ &\left.+2\left(y_2+y_4\dots+y_{n-2}\right)\right] \end{align*} $$n=\frac{x_n-x_0}{h}=\frac{8-0}{2}=4$$ $$I=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2\left(y_2\right)\right]$$ $$y_0=f(x_0)=f(0)=\frac{1}{5}$$ $$y_1=f(x_1)=f(2)=\frac{1}{9}$$ $$y_2=f(x_2)=f(4)=\frac{1}{21}$$ $$y_3=f(x_3)=f(6)=\frac{1}{41}$$ $$y_4=f(x_4)=f(8)=\frac{1}{69}$$ \begin{align*} I&={\scriptstyle \frac{2}{3}\left[\frac{1}{5}+\frac{1}{69}+4\left(\frac{1}{9}+\frac{1}{41}\right)+2\left(\frac{1}{21}\right)\right]}\\ &=0.568 \end{align*}

Hence, answer is (A)

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)

Problem set 45

1. A plane electromagnetic wave is traveling along the positive z-direction. The maximum electric field along the x-direction is 10 V/m. The approximate maximum values of the power per unit area and the magnetic induction $B$, respectively, are
1. $3.3\times 10^{-7} watts/m^2$ and 10 tesla
2. $3.3\times 10^{-7} watts/m^2$ and $3.3\times 10^{-8} $ tesla
3. $0.265 watts/m^2$ and $10$ tesla
4. $0.265 watts/m^2$ and $3.3\times 10^{-8} $ tesla

The power per unit area transported by the electromagnetic waves is given \begin{align*} I&=c\epsilon_0E_0^2\\ &=3\times 10^8\times8.854\times10^{-12}\times10^2\\ &=0.265 watts/m^2 \end{align*}

Magnetic induction \begin{align*} B&=\frac{1}{c}E_0\\ &=\frac{1}{3\times10^{-8}}\times10\\ &=3.3\times 10^{-8} \: tesla \end{align*}

Hence, answer is (D)

2. A particle moves in one dimension in the potential $V=\frac{1}{2}k(t)x^2$, where $k(t)$ is a time dependent parameter. Then $\frac{d}{dt}\left < V\right >$, the rate of change of the expectation value $\left < V\right >$ of the potential energy, is
1. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{k}{2m}\left < xp+px\right >$
2. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{1}{2m}\left < p^2\right >$
3. $\frac{k}{2m}\left < xp+px\right >$
4. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right >$

$$\frac{d\left <\hat A\right >}{dt}=\left < \frac{\partial \hat A}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\hat H,\hat A\right]\right > $$ \begin{align*} &\frac{d\left < V\right >}{dt}={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\frac{p^2}{2m}+V, V\right]\right >}\\ &={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,V\right]\right > }\\ &={\scriptstyle\left < \frac{\partial \left(\frac{1}{2}k(t)x^2\right)}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,\frac{1}{2}k(t)x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < \left[p^2,x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > }\\ &{\scriptstyle+\frac{ik}{4m\hbar}\left < xp\left[p,x\right]+\left[p,x\right]p+p\left[p,x\right]x+\left[p,x\right]xp\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < -2i\hbar xp-2i\hbar px \right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{k}{2m}\left < xp+ px \right > } \end{align*}

3. Consider three inertial frames of reference A, B and C. The frame B moves with a velocity $c/2$ with respect to $A$, and $C$ moves with velocity $c/10$ with respect to B in the same direction. The velocity of C as measured in A is
1. $\frac{3c}{7}$
2. $\frac{4c}{7}$
3. $\frac{c}{7}$
4. $\frac{\sqrt{3}c}{7}$

Let $V$ be the velocity of frame $B$ with respect to frame A, $u_x$ be the velocity of frame C with respect to frame A, and $u_x'$ be the velocity of frame C with respect to frame B, then we have $$u_x'=\frac{u_x-V}{1-u_xV/c^2}$$and $$u_x=\frac{u_x'+V}{1+u_x'V/c^2}$$ \begin{align*} u_x&=\frac{u_x'+V}{1+u_x'V/c^2}\\ &=\frac{\frac{c}{10}+\frac{c}{2}}{1+\frac{c}{10}\frac{c}{2}\frac{1}{c^2}}=\frac{4}{7}c \end{align*}

Hence, answer is (B)

4. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Using relation $$E_{left}\epsilon_{left}\cos{\theta_1}=E_{right}\epsilon_{right}\cos{\theta_2}$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and the vector normal to the yz-plane and $\theta_2$ is the angle between $\vec E_{right}$ and the vector normal to the yz-plane. The vector normal to yz-plane is $\hat i$

Using $\epsilon_{left}:\epsilon_{right}=1:2$ in above equation $$E_{left}\cos{\theta_1}=2E_{right}\cos{\theta_2}\quad--(1)$$ Magnitude of $\vec E_{left}$ is given by $$E_{left}=\sqrt{c^2+c^2+c^2}=\sqrt{3}c$$ Let us find angle $\cos{\theta_1}$ \begin{align*} \cos{\theta}&={\textstyle\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}}\\ &=\frac{c}{\sqrt{3c^2}\sqrt{1}}=\frac{1}{\sqrt{3}} \end{align*} Hence, equation (1) becomes $$\sqrt{3}c\times\frac{1}{\sqrt{3}}=2E_{right}\cos{\theta_2}$$ $$E_{right}\cos{\theta_2}=\frac{c}{2}$$ One cane verify that $E_{right}=c\left(\hat i+2\hat j+2\hat k\right)$ satisfies the above condition. For this vector find $E_{right}$ and $\cos\theta_2$.

Hence, answer is (B)

5. Which of the following transformations $\left(V,\vec A\right)\rightarrow \left(V',\vec A'\right)$ of electrostatic potential $V$ and the vector potential $\vec A$ is a gauge transformation?
1. $\left(V'=V+ax,\vec A'=\vec A+at\hat k\right)$
2. $\left(V'=V+ax,\vec A'=\vec A-at\hat k\right)$
3. $\left(V'=V+ax,\vec A'=\vec A+at\hat i\right)$
4. $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

The general gauge transformation of electrostatic potential $V$ and the vector potential $\vec A$ is given by the equations $$ V'=V-\frac{\partial f}{\partial t}$$ $$\vec A'=\vec A+\nabla f$$ In all options, we have, $V'=V+ax$. This implies $$-\frac{\partial f}{\partial t}=ax$$ Hence, $f=-atx$

Hence, $\nabla f=-at\hat i$

Hence, we have, $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

Hence, answer is (D)

Problem set 44

1. Three polarizers P₁, P₂ and P₃ are arranged as shown in figure. Optical axis of P₁ and P₃ are perpendicular to one another while the axis of P₂ makes an angle $\theta$ with that of P₁. If a beam of ordinary light of intensity I₀ is incident normally on P₁, then the intensity of light emerging from P₃ is :
   
1. 0
2. $\frac{I_0}{8}\sin^2{2\theta}$
3. $\frac{I_0}{2}$
4. $\frac{I_0}{4}\sin^2{2\theta}$
According to Malus law when a beam of polarized light of intensity $I_0$ falls on the polarizer, the intensity of transmitted light is given by $$I=I_0\cos^2\theta$$ where, $\theta$ is the angle between the polarizer axis and polarization direction of light.

When ordinary light falls on the polarizer, then the intensity of light passed through polarizer is one half the intensity of unpolarized light

Hence, for polarizer $P_1$, we have $$I_1=\frac{I_0}{2}$$ For polarizer $P_2$, according to Malus law $$I_2=I_1\cos^2\theta=\frac{I_0}{2}\cos^2\theta$$ For polarizer $P_3$, according to Malus law \begin{align*} I_3&=\frac{I_2}{2}\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\sin^2{(\theta)}\\ &=\frac{I_0}{8}\left(2\cos\theta\sin{(\theta)}\right)^2\\ &=\frac{I_0}{8}\sin^2{2\theta} \end{align*} Hence, answer is (B)

2. There is no infrared absorption for nitrogen molecule because:
1. its polarizability is zero
2. it has no vibrational levels
3. it has no rotational levels
4. its dipole moment is zero
According to selection rule of IR absorption there should be change in dipole moment of the molecule. Nitrogen molecule is homonuclear diatomic molecule. It has no dipole moment and also there is no change in its dipole moment during vibration.

Hence answer is (D)

3. In the first order Stark effect in hydrogen atom the ground state split into :
1. 2 levels
2. 3 levels
3. 4 levels
4. does not split
Splitting of spectral lines in the presence of external electric field is called a Stark effect. The first order correction to the ground state energy of hydrogen atom in the presence of electric $\bar\epsilon$ field applied in z-direction is given by \begin{align*} \Delta E_0&=\int\psi^*_{100}H'\psi_{100}d\tau\\ &=\int\psi^*_{100}(e\epsilon z)\psi_{100}d\tau\\ &=e\epsilon\int z\left|\psi_{100}\right|^2d\tau\\ &=0 \end{align*} Since, integrand is odd function. Hence, there is no splitting in the ground state energy.

Hence, answer is (D)

4. According to Hund's rule, the ground state of Si (atomic number 14) atom is
1. $^1P_1$
2. $^3P_0$
3. $^3P_3$
4. $^3D_3$
The electron configuration of Si is $1s^22s^22p^63s^23p^2$. Hence, there are 2 unpaired electrons in $p$ state. Hence, total spin is $S=\frac{1}{2}+\frac{1}{2}=1$. Multiplicity is $(2S+1)=3$.

For $3p^2$ electrons $m_{l_1}=+2$, $m_{l_2}=+1$, hence $L=2+1=3$

Hence, $J=L-S=0$

term symbol is $^3P_0$

Hence, answer is (B)

5. A source with bandwidth of $10^{-3}\:nm$ centered about $\lambda=500\:nm$ has a coherence length
1. 0.25 m
2. 2.5 $\mu$m
3. 25 cm
4. 2.5 m
Coherence length \begin{align*} l_c&=\frac{\lambda^2}{\Delta\lambda}\\ &=\frac{25\times10^{4}}{10^{-3}}\\ &=25\times10^{7} nm=0.25 m \end{align*}

Problem set 43

1. A positive clamping circuit is one that clamps:
1. The positive extremity of the signal to the zero level
2. The positive extremity of the signal to a positive dc voltage
3. The negative extremity of the signal to the zero level
4. The negative extremity of the signal to a positive dc voltage

A circuit that places either the positive or negative peak of a signal at a desired D.C level is known as a clamping circuit. A clamping circuit introduces (or restores) a D.C level to an A.C signal. Thus a clamping circuit is also known as D.C restorer, or D.C reinserted or a baseline stabilizer. The following are two general types of clamping.

1. Positive clamping occurs when negative peaks raised or clamped to ground or on the zero level i.e., it pushes the signal upwards so that negative peaks fall on the zero level.
2. Negative clamping occurs when positive peaks raised or clamped to ground or on the zero level i.e., it pushes the signal downwards so that the positive peaks fall on the zero level.

In both cases the shape of the original signal has not changed, only there is vertical shift in the signal

Hence, answer is (C).

2. A positive logic NAND gate performs same as the negative logic:
1. XOR gate
2. OR gate
3. AND gate
4. NOR gate
The truth table of positive logic NOR gate is

0	0	1
1	0	0
0	1	0
1	1	0

Now invert all values (as if you place inverters on the two inputs and the one output) to find truth table of negative logic NOR gate:

1	1	0
0	1	1
1	0	1
0	0	1

Which is truth table of positive NAND gate

Hence, answer is (D)

3. If $C=0.1\mu F$, $R=3.25k\Omega$ in a phase shift oscillator feedback circuit, then the frequency of oscillation is:
1. 200 kHz
2. 100 Hz
3. 200 Hz
4. 100 kHz
Frequency of phase shift oscillator is $$f_r=\frac{1}{2\pi RC\sqrt{2N}}$$ where, $R$ is the Resistance in Ohms, $C$ is the Capacitance in Farads, and $N$ is the number of RC stages. In phase shift oscillator feed back circuit there are 3 RC stages. \begin{align*} f_r&={\textstyle \frac{1}{2\times3.14\times3.25\times10^{3}\times10^{-7} \times\sqrt{2\times 3}}}\\ &=200 Hz \end{align*}

Hence, answer is (C)

4. Load regulation is determined by :
1. Changes in load current and output voltage
2. Changes in load current and input voltage
3. Changes in load resistance and input voltage
4. Changes in Zener current and load current
Load regulation indicates the change in output voltage due to the change in load current. There is another type of voltage regulation, called line regulation and indicates the change in output voltage due to the change in input voltage.


Hence, answer is (A)

5. A carrier is simultaneously modulated by two sine waves with modulation indices of 0.3 and 0.4; then the total modulation index is :
1. 1
2. 0.1
3. 0.5
4. 0.35
$$m_1 = 0.3, m_2 = 0.4$$ $$m_t=\sqrt{m_1^2+m_2^2}=0.5$$

Hence, answer is (C)

Problem set 42

1. The input signal for the equivalent circuit shown below can have a frequency between 10 Hz and 50 kHz, then the value of the coupling capacitor is:
   
1. $1\mu F$
2. $10 pF$
3. $1 pF$
4. $10\mu F$
For coupling capacitive reactance $X_C$ must be low. Since, $X_C=\frac{1}{2\pi fC}$, $X_C$ is inversely proportional to frequency $f$. Hence, value of $C$ is decided by the lowest frequency of operation. Now, in a circuit, for a coupling, the value of $C$ should be such that $X_C=\frac{\text{total resistance of circuit}}{10}$. $${\scriptstyle \text{total resistance of circuit}=R=12k\Omega||24k\Omega||3k\Omega||6k\Omega}$$ $$\frac{1}{R}=\frac{1}{12}+\frac{1}{24}+\frac{1}{3}+\frac{1}{6}=\frac{15}{24}$$ $$R=\frac{24}{15}$$ $$X_C=\frac{R}{10}=\frac{24}{150}k\Omega$$ \begin{align*} C&=\frac{1}{2\pi fX_C}\\ &=\frac{1}{2\times3.14\times10\times\frac{24}{150}\times10^3}\\ &=100\mu F \end{align*} However, if we take ${\scriptstyle X_C=\text{total resistance of circuit}=R=\frac{24}{15}}$, we get, \begin{align*} C&=\frac{1}{2\pi fX_C}\\ &=\frac{1}{2\times3.14\times10\times\frac{24}{15}\times10^3}\\ &=10\mu F \end{align*} Hence, answer is (D)
2. In a 3-input OP-AMP summing amplifier shown below, the output voltage $(v_0)$ is
   
1. -3 V
2. +3 V
3. +6 V
4. -9 V
Applying KCL law to inverting node $$\frac{v_1}{R_1}+\frac{v_2}{R_2}+\frac{v_3}{R_3}+\frac{v_0}{R_f}=0$$ \begin{align*} v_0&=-R_f\left(\frac{v_1}{R_1}+\frac{v_2}{R_2}+\frac{v_3}{R_3}\right)\\ &={\scriptstyle -10^3\left(\frac{2}{3\times10^3}+\frac{3}{3\times10^3}+\frac{4}{3\times10^3}\right)=-3}\\ \end{align*} Hence, answer is (A)
3. In the circuit given below what is the approximate ac voltage across the output resistor:
   
1. 15 mV
2. 150 mV
3. 15 $\mu$V
4. 15 V
This is a single stage amplifier. $$v_0=A_vv_{in}$$ where, $A_v$ is ac voltage gain given by $A_v= $
4. The input impedance $(Z_{in(total)})$ of the common-emitter amplifier given below is:
   
1. $5k\Omega$
2. $4k\Omega$
3. $2k\Omega$
4. $20k\Omega$
As $\beta=200$, the value of the transistor base impedance $=\beta(r_e+R_E)=\beta r_e$, since $R_E=0$.
transistor base impedance $=\beta r_e=200\times25=5k\Omega$ $$Z_{in(total)}=20||\beta r_e$$ \begin{align*} \frac{1}{Z_{in(total)}}&=\frac{1}{20}+\frac{1}{\beta r_e}\\ &=\frac{1}{20}+\frac{1}{5}=\frac{1}{4} \end{align*} $$Z_{in(total)}=4k\Omega$$ Hence, answer is (B)
5. The load voltage in a Zener circuit shown below with $V_z=15V$ is approximately
   
1. 15 V
2. 10 V
3. 14.3 V
4. 15.7 V
$$V_L=V_Z+V_{BE}=15-0.7=14.3 V$$ $V_{BE}$ negative because it has polarity opposite to that of $V_Z$.
Hence, answer is (C)

Problem set 41

1. In the op-amp circuit shown in the figure, $V_i$ is a sinusoidal input signal of frequency $10 Hz$ and $V_0$ is the output signal.
   
   The magnitude of the gain and the phase shift, respectively, close to values
1. $5\sqrt{2}$ and $\pi/2$
2. $5\sqrt{2}$ and $-\pi/2$
3. 10 and zero
4. 10 and $\pi$
Here $R_1=1k=10^3\Omega$, $R_f=10k=10^4\Omega$, $C=0.01\mu F=0.01\times10^{-6}F=\times10^{-8}F$, $f=10Hz$. At inverting terminal current is zero. Applying Kirchoff's current law at inverting terminal, we have $$\frac{V_i}{R_1}+\frac{V_0}{R_f||Z_C}=0$$ $$\frac{V_0}{R_f||Z_C}=-\frac{V_i}{R_1}$$ \begin{align*} \frac{V_0}{V_i}&=-\frac{R_f||Z_C}{R_1}\\ &=-\frac{R_fZ_C}{R_1(R_f+Z_C)} \end{align*} $$Z_C=\frac{X_C}{j}\quad\text{where }j=\sqrt{-1} $$ \begin{align*} X_C&=\frac{1}{\omega C}=\frac{1}{2\pi f C}\\ &=\frac{1}{2\pi\times10\times 10^{-8}}=\frac{10^{7}}{2\pi} \end{align*} $$Z_C=\frac{10^{7}}{j2\pi}$$ \begin{align*} \frac{V_0}{V_i}&=-\frac{10^4\times \frac{10^{7}}{j2\pi}}{10^3\left(10^4+\frac{10^{7}}{j2\pi}\right)}\\ &=-\frac{10^1}{\left(j2\pi10^{-4}+1\right)}\\ &=-\frac{10\left(1-j2\pi10^{-4}\right)}{\left(-4\pi^210^{-8}+1\right)}\\ \end{align*} $$\left|\frac{V_0}{V_i}\right|\simeq10$$ For parallel RC-circuit phase shift is \begin{align*} \phi&=-\tan^{-1}{\left(\frac{R}{X_C}\right)}\\ &=-\tan^{-1}{\left(\frac{{2\pi}\times10^4}{10^{7}}\right)}\\ &\simeq-\tan^{-1}{\left(0\right)}=\pi \end{align*} Hence, answer is (D)
2. The logic circuit shown in the figure below

implements the Boolean expression
1. $y=\overline{A\cdot B}$
2. $y=\bar{A}\cdot \bar{B}$
3. $y=A\cdot B$
4. $y=A+ B$
Exor gate has the output high if the inputs are not alike otherwise the output is low. Hence, output of two Exor gates will be $\bar{A}$ and $\bar{B}$. Hence, $y=\bar{A}+\bar{B}=\overline{A\cdot B}$ (DeMorgan’s Law)
Hence, answer is (A)
3. A time varying signal $V_{in}$ is fed to an op-amp circuit with output signal $V_0$ as shown in the figure below.
   
   The circuit implements a
1. high pass filter with cutoff frequency 16Hz
2. high pass filter with cutoff frequency 100 Hz
3. low pass filter with cutoff frequency 16Hz
4. low pass filter with cutoff frequency 100 Hz
In the circuit lower op-amp is integrator and integrator is a low pass-filter. OR In the lower op-amp non-inverting terminal is grounded. Hence, inverting terminal is virtual ground. Hence, capacitor is grounded. Hence, the RC combination of (10k) and $1\mu F$ acts as a low pass filter. Its cutoff frequency is given by $f=\frac{1}{2\pi RC}$. \begin{align*} f&=\frac{1}{2\times3.14\times 10\times10^3\times1\times10^{-6}}\\ &=15.91=16 Hz \end{align*} Hence, answer is (C).
4. In the following circuit the current through the load resistance is:
   
1. 10 mA
2. 1 mA
3. 5 mA
4. 0.5 mA
Voltage across load resistance is equal to voltage across zener diode. Hence, $I=\frac{10}{2\times10^3}=5 mA$.
Hence, answer is (C)
5. In the following clipping circuit, the clipping level is:
   
1. + 25 V
2. - 25 V
3. - 5 V
4. + 5 V
Due to voltage divider arrangement, the voltage drop across lower $50\Omega$ resistance is $V_D=+5 V$. Hence, clipping level is $V_R+V_D=5+0.6=5.6 V$
Hence, answer is (D)

Problem set 40

1. If superconducting lead has critical temperature of 7.26 K at zero magnetic field and a critical field of $8\times10^5 A/m$ at 0K, the critical field at 5K is :
1. $6.3\times 10^5 A/m$
2. $2.5\times 10^5 A/m$
3. $4.2\times 10^5 A/m$
4. $1.5\times 10^5 A/m$
$H_{c_0}=8\times10^5A/M$ critical field at 0 K
$T=5K$ Temperature at which critical field of lead is to be found outside
$T_c=7.26 K$ Critical temperature of lead
\begin{align*} H_{c_T}&=H_{c_0}\left(1-\left(\frac{T}{T_c}\right)^2\right)\\& =8\times10^5\left(1-\left(\frac{5}{7.26}\right)^2\right)\\ &=4.2\times10^5 A/m \end{align*}
2. 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)
3. 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)
4. 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 it provides an output that is proportional to dose.
Hence, answer is (B)
5. 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.

Problem set 39

1. A state of a system with spherically symmetric potential has zero uncertainty in simultaneous measurement of operator $L_x$ and $L_y$. Which of the following statement is true?
1. The state must be $l=0$ state
2. Such a state can never exist as operators $L_x$ and $L_y$ do not commute
3. The state has $l=1$ with $m=0$
4. The state cannot be an eigenstate of $L^2$ operator
\begin{align*} L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\ &=l(l+1)\hbar^2\psi_{nlm} \end{align*} $$\left < L_z \right > =m\hbar, \left < L_z^2 \right > =m^2\hbar^2$$ $$\text{ and } \left < L^2 \right > =l(l+1)\hbar^2$$ The state $\psi_{nlm}$ is symmetric in $x$ and $y$, so \begin{align*} \left < L_x^2 \right >&=\left < L_y^2 \right > \\ &=\frac{\left(\left < L^2 \right >-\left < L_z^2 \right >\right)}{2}\\ &=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2} \end{align*} It can be verified that $\left < L_x \right >=\left < L_y \right >=0$. Using, $\Delta L_i=\sqrt{\left < L_i^2 \right >-\left < L_i \right >^2}$, we have $$\Delta L_x=\Delta L_y=\sqrt{\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}}$$ $$\Delta L_x\Delta L_y=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}$$ From this relation one can see that uncertainty relation will be zero when $l=0$ and henc, $m=0$.
Hence, answer is (A)
2. The wave function for identical fermions is antisymmetric under particle exchange. Which of the following is a consequence of this property?
1. Heisenberg's uncertainty principle
2. Bohr correspondence principle
3. Bose-Einstein condensation
4. Pauli exclusion principle
(D) Pauli exclusion principle
3. The spin part of two electron wave function is described as a triplet state. The space part of the wave function is given by ($\psi_1$ and $\psi_2$ are two different functions):
1. $\psi_1(r_1)\psi_2(r_2)$
2. $\psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)$
3. $\psi_1(r_1)\psi_2(r_2)+\psi_2(r_1)\psi_1(r_2)$
4. $\psi_1(r_1)\psi_1(r_2)+\psi_2(r_1)\psi_1(r_2)$
Triplet state means $S=s_1+s_2=1$ and $M_s=-1,0,+1$. For $M_s=-1$ we must have $m_{s1}=-\frac{1}{2}$ and $m_{s2}=-\frac{1}{2}$
For $M_s=0$ we must have either $m_{s1}=\frac{1}{2}$ and $m_{s2}=-\frac{1}{2}$ or $m_{s1}=-\frac{1}{2}$ and $m_{s2}=\frac{1}{2}$
For $M_s=+1$ we must have $m_{s1}=+\frac{1}{2}$ and $m_{s2}=+\frac{1}{2}$
Hence, we have three possible spin eigenfunctions $$\phi\left(+\frac{1}{2},+\frac{1}{2}\right)$$ $$\frac{1}{\sqrt{2}}\left[\phi\left(+\frac{1}{2},-\frac{1}{2}\right)+\phi\left(-\frac{1}{2},+\frac{1}{2}\right)\right]$$ $$\phi\left(-\frac{1}{2},-\frac{1}{2}\right)$$ All these functions are symmetric. However, total wavefunction must be antisymmetric. Hence, space part must be antisymmetric.
Hence, answer is (B) i.e. $\psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)$
4. A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from $\psi_1$ to $\psi_2$ is forbidden if $\psi_1$ and $\psi_2$
1. have opposite parity
2. are both spherically symmetrical
3. are orthogonal to each other
4. are zero at the center of the atomic nucleus
Selection rules:

1.	Principal quantum number	:	$\Delta n=$ anything
2.	Orbital angular momentum	:	$\Delta l = \pm1$
3.	Magnetic quantum number	:	$\Delta m_l = 0, \pm1$
4.	Spin	:	$\Delta s = 0$
5.	Total angular momentum	:	$\Delta j = 0, \pm 1,{\scriptstyle \text{ but} j = 0 \nrightarrow j= 0}$

A. FALSE: It's not related to the selection rules.
B. TRUE: If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. $l = 0 \rightarrow l = 0$ is forbidden as $\Delta l=0$ in this case.
C. FALSE: In any transition, eigenstates should always be mutually orthogonal.
D. FALSE: Eigenstates zero at the center $\Rightarrow l> 0$ could change, for example from 3 to 2. This is allowed.
Hence, answer is (B)
5. 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

Problem set 38

1. Merit of a solar cell or fill factor is
1. $I_mV_m/I_{SC}V_{OC}$
2. $I_mV_{OC}/I_{SC}V_{m}$
3. $I_{SC}V_{OC}/I_{m}V_{m}$
4. $I_{SC}V_{m}/I_{m}V_{OC}$
(A) $I_mV_m/I_{SC}V_{OC}$
2. The resonant frequency of a Heartly oscillator with $L_1=12\mu H$, $L_2=8\mu H$ and $C=1000PF$ is:
1. 1.12 MHz
2. 11.2 MHz
3. 11.2 kHz
4. 112 kHz
$$f_r=\frac{1}{2\pi\sqrt{L_TC}}$$ where, ${\scriptstyle L_T=L_1+L_2=12+8=20\mu H=20\times10^{-6}H}$, ${\scriptstyle C=1000PF=1000\times10^{-12}=10^{-9}F}$ \begin{align*} f_r&=\frac{1}{2\times3.14\sqrt{20\times10^{-6}\times10^{-9}}}\\ &=1.12 MHz \end{align*} Hence, answer is (A)
3. In an open loop differential operational amplifier having gain of $A=2\times10^5$ receives inputs as at non-inverting terminal $5\mu V$ and at inverting terminal $-7\mu V$, then the output is:
1. 2.4 V
2. 0.24 V
3. 2.4 mV
4. 2.4 $\mu$V
\begin{align*} V_o&=-A(V_2-V_1)\\ &=2\times10^5(5-(-7))\times10^{-6}\\ &=2.4V \end{align*} Hence, answer is (A)
4. The number of Full-adders and Half-adders required to addd 16-bit numbers is
1. 1 HA and 15 FA
2. 8 HA and 08 FA
3. 16 HA and 0 FA
4. 4 HA and 12 FA
(A) 1 HA and 15 FA
5. Sum of all the three inputs will appear as output from:
1. A 3-input NAND gate followed by an inverter
2. A 3-input XOR gate followed by an inverter
3. A 3-input NOR gate followed by an inverter
4. An inverter followed by 3-input NOR gate
Three input NOR gate has output $Z=\bar{X+Y+W}$. Hence, this followed by an inverter will give sum of all the three inputs.
Hence, answer is (A)

Problem Set 37

1. The LS configurations of the ground state of $^{12}Mg$, $^{13}Al$, $^{17}Cl$ and $^{18}Ar$ are, respectively,
1. $^3S_1$, $^2P_{1/2}$, $^2P_{1/2}$ and $^1S_{0}$
2. $^3S_1$, $^2P_{3/2}$, $^2P_{3/2}$ and $^3S_{1}$
3. $^1S_0$, $^2P_{1/2}$, $^2P_{3/2}$ and $^1S_{0}$
4. $^1S_0$, $^2P_{3/2}$, $^2P_{1/2}$ and $^3S_{1}$

Electronic configuration of $^{12}Mg$ is $1s^22s^22p^63s^2$

There are two 3s electrons with $s_1=\frac{1}{2}$ and $s_2=\frac{1}{2}$. Hence, $S=1$ and $2S+1=3$

For two 3s electrons each with $l=0$, $m_l=0$. Hence, $L=0$

$J=L+S=0$. Hence, term symbol for $^{12}Mg$ is $^1S_0$

For, $^{13}Al$, electronic configuration is $1s^22s^22p^63s^23p^1$

$S=\frac{1}{2}$, $2S+1=2$, $L=1$, $J=L-S=1-\frac{1}{2}=\frac{1}{2}$ (less than half-filled)

Hence, term symbol for $^{13}Al$ is $^2P_{1/2}$

Hence, answer is (C).

2. The frequency dependent dielectric constant of a material is given by $$\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}$$ where $A$ is a positive constant, $\omega_0$ the resonant frequency and $\gamma$ the damping coefficient. For an electromagnetic wave of angular frequency $\omega < < \omega_0$, which of the following is true? (Assume that $\frac{\gamma}{\omega_0} < < 1$.)
1. There is negligible absorption of the wave
2. The wave propagation is highly dispersive
3. There is strong absorption of the electromagnetic wave
4. The group velocity and the phase velocity will have opposite sign
\begin{align*} &\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2+i\omega\gamma\right)\left(\omega_0^2-\omega^2-i\omega\gamma\right)}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}+i\frac{A\omega\gamma}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ \end{align*} Fro $\omega < < \omega_0$, imaginary part of $\epsilon(\omega)$ is negligibly small. Hence, there is negligible absorption of the wave.

Hence, answer is (A)

3. The state diagram corresponding to the following circuit is
   

In options, ⓪ and ① represent states of Flip-Flop i.e. output $A$. Suppose flip-flop is in state ⓪ i.e. $A=0$. If the $x, y$ inputs are $0,0$ then output $A$ will become 1 i.e. state will change from ⓪ to ①

Now, suppose the flip-flop is in state ⓪ i.e. $A=0$. If the $x, y$ inputs are $1,1$ then output $A$ will remain 0 i.e. the state remains same ⓪.

Hence, option (D) is correct

4. Consider a random walker on a square lattice. At each step the walker moves to a nearest neighbour site with equal probability for each of the four sites. The walker starts at the origin and takes 3 steps. The probability that during this walk no site is visited more than once is
1. 12/27
2. 27/64
3. 3/8
4. 9/16

During first step of random walk the probability that during this walk no site is visited more than once is 1

During second step of random walk the probability that during this walk no site is visited more than once is 3/4, because there are 3 sites which are unvisited and each site has probability 1/4

Similarly, during third step of random walk the probability is 3/4

Hence, the probability that during first, second and third step, no site is visited more than once is $$p=1\times\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$$

Hence, answer is (D)

5. A canonical transformation $(p,q)\rightarrow(P,Q)$ is performed on the Hamiltonian $H=\frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$ via the generating function $F=\frac{1}{2}m\omega q^2\cot{Q}$. If $Q(0)=0$, which of the following graphs shows schematically the dependence of $Q(t)$ on $t$?
1. 
2. 

4. 
$H=\frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$, $F=\frac{1}{2}m\omega q^2\cot{Q}$ $$H'=H-\frac{\partial F}{\partial t}=H$$ For the generating function $F$ we have $\frac{\partial F}{\partial q}=p$ and $\frac{\partial F}{\partial Q}=-P$

$\frac{\partial F}{\partial q}=p$ gives $$m\omega q\cot{Q}=p$$

$\frac{\partial F}{\partial Q}=-P$ gives $$-\frac{1}{2}m\omega q^2 \text{ cosec}^2Q=-P$$ $$q^2=\frac{2P}{m\omega \text{ cosec}^2Q}$$ \begin{align*} H'&={\scriptstyle\frac{1}{2m}m^2\omega^2q^2\cot^2Q+\frac{1}{2}m\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}}\\ &={\scriptstyle\frac{1}{2m}m^2\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}\cot^2Q+\frac{1}{2}m\omega^2\frac{2P}{m\omega \text{ cosec}^2Q}}\\ &=P\omega\cos^2Q+P\omega\sin^2Q\\ &=P\omega \end{align*} Using canonical equation $\dot Q=\frac{\partial H'}{\partial P}$ we have $$\frac{dQ}{dt}=\omega$$ $$Q=\omega t+c$$ where $c$ in integrating constant. But at $t=0$, $Q(t)=0$, hence, $c=0$ $$Q=\omega t$$ Which is equation of straight line.

Hence, answer is (D)

Problem set 36

1. For a finite square well potential in one dimension:
1. It is possible that no bound state exits
2. There is always at least one bound state
3. Bound states have degeneracy = 2
4. Energy levels of bound states are equally spaced
(B) There is always at least one bound state
2. A particle with spin $\frac{1}{2}$ is in state with eigenstate of $S_z$. Then the expectation values of $S_x$, $S_x^2$ in this state are given by:
1. $-\frac{\hbar}{2}$, $\frac{1}{4}\hbar$
2. $0$, $\frac{3}{4}\hbar^2$
3. $\frac{\hbar}{2}$, $\frac{3}{4}\hbar^2$
4. $0$, $\frac{1}{4}\hbar^2$
For spin $\frac{1}{2}$ particle the eigenstates are given by $\chi_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\chi_2=\begin{bmatrix}0\\1\end{bmatrix}$. The spin components $\hat S_x$, $\hat S_y$ and $\hat S_z$ are given by $$\hat S_x=\frac{\hbar}{2}\sigma_x=\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}$$ $$\hat S_y=\frac{\hbar}{2}\sigma_y=\frac{\hbar}{2}\begin{bmatrix}0&-i\\i&0\end{bmatrix}$$ $$\hat S_z=\frac{\hbar}{2}\sigma_z=\frac{\hbar}{2}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ \begin{align*} \left < \hat S_x\right > &=\left < \chi_1\left|\hat S_x\right|\chi_1\right > \\ &=\begin{bmatrix}1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ &=0 \end{align*} \begin{align*} \left < \hat S_x^2\right > &=\left < \chi_1\left|\hat S_x\hat S_x\right|\chi_1\right > \\ &{\scriptstyle=\begin{bmatrix}1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\frac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}}\\ &=\frac{1}{4}\hbar^2 \end{align*} Hence, answer is (D)
3. The differential cross-section for a central potential is equal to
1. $f(\theta,\phi)$
2. $f^*(\theta,\phi)$
3. $f^*(\theta,\phi)f(\theta,\phi)$
4. $|f(\theta,\phi)|$
where asymptotic form of the wave function of the relative motion is given by: $$A\left[e^{ikz}+\frac{f(\theta,\phi)}{r}e^{ikr}\right]$$
Differential cross-section is given by $\frac{d\sigma}{d\Omega}=|f(\theta,\phi)|^2$. Hence, answer is (C).
4. The de Broglie wavelength of a helium atom at 300 K is 0.06 $A^o$. The de Broglie wavelength of neon atom (5 times heavier than helium) at 600 K will be:
1. 6 Å
2. 0.06 Å
3. $0.06\times\sqrt{10}$Å
4. $\frac{0.06}{\sqrt{10}}$Å
Thermal de Broglie wavelength is given by $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$. In the nonrelativistic case the effective kinetic energy of free particles is $E = \pi k_B T $. $$\lambda=\frac{h}{\sqrt{2\pi mk_BT}}$$ $$\lambda_{He}\propto\frac{1}{\sqrt{m_{He}T}}$$ and$$\lambda_{Ne}\propto\frac{1}{\sqrt{m_{Ne}T}}$$ $$\frac{\lambda_{Ne}}{\lambda_{He}}=\frac{\sqrt{m_{He}T_{He}}}{\sqrt{m_{Ne}T_{Ne}}}$$ $$\lambda_{Ne}=\sqrt{\frac{m_{He}T_{He}}{m_{Ne}T_{Ne}}}\lambda_{He}$$ $\lambda_{Ne}=\sqrt{\frac{1\times 300}{5\times600}}0.6=\frac{0.6}{\sqrt{10}}$Å
Hence, answer is (D)
5. If a charged particle $q$ moves along a circle of radius $r=100mm$ in a uniform magnetic field $B=10mT$, then the period of revolution of the particle $(m_p=1.67\times10^{-27}kg, q=1.6\times10^{-19}C)$
1. 6.55 ms
2. 6.55 $\mu$s
3. 6.55 ns
4. 3$\mu$s
For circular motion radial force is equal to force due to magnetic field. $$\frac{mv^2}{r}=qvB$$ $$T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$$ \begin{align*} T&=\frac{2\times3.14\times1.67\times10^{-27}}{1.6\times10^{-19}\times10\times10^{-3}}\\ &=6.55\mu s \end{align*}

Problem set 35

1. The energy of formation of a vacancy in copper is $1eV$. The number of vacancies per mole below its melting point $1356^oK$ is:
1. $1.15\times10^{20}$
2. $4\times10^{20}$
3. $2\times10^{20}$
4. $3.30\times10^{20}$
$$\frac{n}{N}=e^{\left(-\frac{E}{k_BT}\right)}$$ $$E=-k_BT\ln{\frac{n}{N}}$$ \begin{align*} E&={\scriptstyle-1.38\times10^{-23}\times1356\times\ln{\frac{1}{6.022140857\times10^{23}}}} &=-6.404} \end{align*}
2. A system is known to be in a state described by the wave function $$\psi(\theta,\phi)=\frac{1}{\sqrt{30}}\left[5Y_4^0+5Y_6^0+25Y_6^3\right]$$ where $Y_l^m$ are spherical harmonics. The probability of finding the system in a state with $m=0$ is :
1. Zero
2. $\frac{6}{\sqrt{30}}$
3. $\frac{6}{30}$
4. $\frac{13}{15}$
\begin{align*} |\psi|^2&=\frac{1}{30}\left[25\left|Y_4^0\right|^2+\left|Y_6^0\right|^2\right]\\ &=\frac{26}{30}=\frac{13}{15} \end{align*} Hence, answer is (D)
3. For attractive one-dimensional delta function potential situated at $x=0$, the wave function of the bound state is given by:
1. $\psi(x)=e^{-\alpha x}$
2. $\psi(x)=e^{-\alpha |x|}$
3. $\psi(x)=e^{-\alpha x^2}$
4. $\psi(x)=\sin{\alpha x}$
The attractive delta function is given by $$V(x)=-a\delta(x)$$ The $\delta$-function is infinite at $x=0$ and zero elsewhere. For bound state total energy $E$ of the particle must be less than the potential energy $V(X)$. Hence, $E$ must be negative where $V(x)=0$. The Schrodinger equation at these points will be $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi\quad\text{ or }\frac{d^2\psi}{dx^2}=\alpha^2\psi$$ where, $\alpha=\sqrt{-\frac{2mE}{\hbar^2}}$. The general solution of above equation can be written as $$\psi(x)=Ae^{-\alpha x}+Be^{\alpha x}$$ For $x<0$, the term, $Ae^{-\alpha x}\rightarrow\infty$ as $x\rightarrow-\infty$, and for $x>0$, the term $Ae^{\alpha x}\rightarrow\infty$ as $x\rightarrow\infty$. Hence, we will have to write separate solutions for these two regions. $$\psi_-(x)=Be^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ At $x=0$ these two solutions must be continous. Hence, we get $A=B$. $$\psi_-(x)=Ae^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ Thus, $$\psi(x)=Ae^{-\alpha |x|}\quad\text{for all values of } x$$ Hence, answer is (B)
4. A one-dimensional simple harmonic oscillator with generalized coordinate $q$ is subjected to an extra additional potential energy of the form $$V(t)=q^2t+q\dot qt^2$$ The Lagrange's equation of the oscillator due to the extra potential will contain:
1. an extra term proportional to $t$
2. an extra term proportional to $t^2$
3. an extra term proportional to $(t+t^2)$
4. no extra term
$$L=\frac{1}{2}m\dot q^2-\frac{1}{2}kq^2-q^2t-q\dot qt^2$$ $$\frac{\partial L}{\partial \dot q}=m\dot q-qt^2$$ $$\frac{\partial L}{\partial q}=-kq-2qt-\dot qt^2$$ \begin{align*} &{\scriptstyle\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=\frac{d}{dt}\left(m\dot q-qt^2\right)-\left(-kq-2qt-\dot qt^2\right)}\\ &=m\ddot q-\dot qt^2-2qt+kq+2qt+\dot qt^2\\ &=m\ddot q+kq=0 \end{align*} Hence, there is no change in Lagrange's equation of motion. Hence, answer is (D)
5. The variational method in perturbation theory, when applied to obtain the value of the ground state energy:
1. Always gives exact ground state energy
2. gives energy value lower than the exact ground state energy
3. gives energy value which is sometimes higher than or sometimes lower than the exact ground state energy
4. gives energy value higher than or equal to the exact ground state energy
Variational method gives energy value higher than or equal to the exact ground state energy. Hence, answer is (D).

Problem set 34

1. The second neighbour distance in a simple cubic system having lattice constant $a$ is:
1. $\sqrt{2}a$
2. $a$
3. $\frac{\sqrt{3}}{2}a$
4. $\sqrt{3}a$
Second neighbours are present at the digonal position of the face. Hence, second neighbour distance is $\sqrt{2}a$.
Hence, answer is (A)
2. If the Debye characteristic frequency for copper is $6.55\times10^{12}Hz$, the Debye temperature for copper is :
1. $200K$
2. $314.5K$
3. $405K$
4. $150K$
\begin{align*} \theta_D&=\frac{h\nu_D}{k}\\ &={\scriptstyle\frac{6.62607004\times10^{-34}\times6.55\times10^{12}}{1.38\times10^{-23}}}\\ &=314.5K \end{align*}
3. If the concentration of Schotty defects in a fcc crystal is 1 in $10^{10}$ at $300^oK$, the energy of formation in eV of Schottky defects will be
1. 0.60
2. 0.50
3. 0.40
4. 1.0
$$n=Ne^{\left(-\frac{E_{schottky}}{k_BT}\right)}$$ Here, $n$ is number of defects defects, $N$ is number of atoms. $$\frac{n}{N}=e^{\left(-\frac{E_{schottky}}{k_BT}\right)}$$ $$E_{schottky}=-k_BT\ln{\frac{n}{N}}$$ $$\frac{n}{N}=\frac{1}{10^{10}}$$ \begin{align*} &E_{schottky}={\scriptstyle-1.38\times10^{-23}\times300\ln{\frac{1}{10^{10}}}J}\\ &{\scriptstyle=-\frac{1.38\times10^{-23}\times300}{1.6\times10^{-19}}(-10\ln{10}) eV}\\ &=0.6 eV \end{align*}
4. Van der Waals attractive interaction between inert gas atoms varies with interatomic separation $(R)$ as :
1. $-\frac{1}{R^2}$
2. $-\frac{1}{R^3}$
3. $-\frac{1}{R^6}$
4. $-\frac{1}{R^{12}}$
(C)$-\frac{1}{R^6}$
5. Metallic nickel crystallizes in fcc-type structure with lattice constant $a$. The interplanar distance between the diffracting planes (220) is this material is :
1. $\frac{a}{2\sqrt{2}}$
2. $\frac{a}{2}$
3. $\frac{\sqrt{3}}{2}a$
4. $\frac{a}{\sqrt{2}}$
\begin{align*} d_{hkl}&=\frac{a}{\sqrt{h^2+k^2+l^2}}\\ &=\frac{a}{\sqrt{2^2+2^2+0^2}}=\frac{a}{2\sqrt{2}} \end{align*} Hence, answer is (A)

Problem set 33

1. The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc structure, in which the volume of the unit cell is $a^3$, is
1. $\left(\frac{12\pi^2}{a^3}\right)^{1/3}$
2. $\left(\frac{3\pi^2}{a^3}\right)^{1/3}$
3. $\left(\frac{\pi^2}{a^3}\right)^{1/3}$
4. $\left(\frac{1}{a}\right)^{1/3}$
The radius of Fermi sphere is given by $k_f=\left(\frac{3\pi^2N}{V}\right)^{1/3}$. For FCC structure $\frac{N}{V}=\frac{4}{a^3}$. Hence, $k_f\left(\frac{12\pi^2}{a^3}\right)^{1/3}$. Hence, answer is (A).
2. 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. However, nucleus is not a point charge. There is a charge distribution inside the nucleus. Due to which there is change in angular dependence of scattering. From this information one can estimate size of nucleus.
Hence, answer is (C)
3. If the critical magnetic field for aluminium is $7.9\times 10^3 A/m$, the critical current which can flow through long thin superconducting wire of aluminium of diameter $1\times10^{-3}m$ is
1. 5.1 A
2. 1.6 A
3. 4.0 A
4. 2.48 A
Critical current is given by \begin{align*} I_c&=2\pi rH_c\\ &=2\times3.14*0.5\times10^{-3}\times7.9\times 10^3\\ &=2.48 A \end{align*} Hence, answer is (D)
4. An elemental dielectric has a dielectric constant $\epsilon=12$ and it contains $5\times10^{18} atoms/m^3$. Its electronic polarizability in uints of $FM^2$ is
1. $4.17\times10^{-30}$
2. $8.34\times10^{-30}$
3. $6.32\times10^{-30}$
4. $12.64\times10^{-30}$
The Clausius Mossotti relation is $$\frac{\epsilon_r-1}{\epsilon_r+2}=\frac{N\alpha_e}{3\epsilon_0}$$ \begin{align*} \alpha_e&=\frac{\epsilon_r-1}{\epsilon_r+2}\frac{3\epsilon_0}{N}\\ &=\frac{12-1}{12+2}\frac{3\times8.85\times10^{-12Fm^{-1}}}{5\times10^{18} atoms/m^3}\\ &=4.17\times10^{-30}FM^2\\ \end{align*} Hence, answer is (A)
5. Mangetite $(Fe_3O_4)$ has a cubic structure with a lattice constant of $8.4A^o$. The saturation magnetization in this material in units of A/m is :
1. $6.2\times10^5$
2. $6.2\times10^6$
3. $12.4\times10^6$
4. $12.4\times10^5$
As each $O$ has an oxidation number of $-2$ and there are 4 of them in the formula, the 3 Fe atoms must together have a charge of $+8$. This is consistent with their being two $Fe^{3+}$ ions and one $Fe^{2+}$ per formula unit. The ratio of $Fe^{3+}$ ions to $Fe^{2+}$ is 2 to 1. There are 8 $Fe_3O_4$ units per unit cell. Hence, there are 24 $Fe$ atoms in unit cell, out of which 16 are $Fe^{3+}$ and 8 are $Fe^{2+}$. Out of the 16 $Fe^{3+}$ ions 8 $Fe^{3+}$ ions are at octahedral position and 8 $Fe^{3+}$ ions are at tetrahedral position. The ions in tetrahedral sites oppose the applied magnetic field and the ions at octahedral positions reinforce the field. Consequently, $Fe^{3+}$ ions neutralizes the $Fe^{3+}$ at octahedral positions. In each $Fe^{3+}$ there are 5 unpaired $3d$ electrons and in each $Fe^{2+}$ there are 4 unpaired $3d$ electrons. Each unpaired electron has magnetic moment one Bohr magneton $\mu_B=9.27\times10^{-24}Am^2$. Hence, each $Fe^{2+}$ ion has 4$\mu_B$ magnetic moment. As, unit cell contains 8 $Fe^{2+}$ ions, total magnetic moment is $32\mu_B$.
Hence, saturation magnetization \begin{align*} M&=\frac{32\mu_B}{a^3}\\ &=\frac{32\times9.27\times10^{-24}}{\left(8.4\times10^{-10}\right)^3}\\ &=5.1\times10^5A/m \end{align*}