Problem set 66

1. Consider a configuration of a system of 10 distinguishable particles in which there are 3 particles in state 1, 3 particles in state 2 and 4 particles in state 3. The total number of microstates is :
1. 4200
2. 864
3. 102060
4. 360

Number of ways of arranging $N$ distinguishable particles into groups of $N_j$ each, such that $N_1$ in state 1, $N_2$ in state 2, etc. and $N_j$ in the $j^{th}$ state are given by $$W(N)=\frac{N!}{N_1!N_2!\dots N_j!}$$ $$W(10)=\frac{10!}{3!\times 3!\times 4!}=4200$$

Hence, answer is (A)

2. In the density matrix representation, the condition for a pure state is:
1. $\hat\rho^2=\hat\rho$
2. $\hat\rho^2=\hat I$
3. $T_r\hat\rho^2=T_r\hat\rho$
4. $T_r\hat\rho=1$

The Density matrix for the pure state $|\psi>$ is given by $$\rho=|\psi><\psi|$$ This density matrix has the following properties:

1. $\hat\rho^2=\hat\rho$ projector
2. $\hat\rho^\dagger=\hat\rho$ hermiticity
3. $T_r\hat\rho=1$ normalization
4. $\hat\rho\geq 0$ positivity

Hence, answer is (A)

3. 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)

4. Analysis of a X-ray diffraction pattern of a material crystallizing in a fcc-type structure gives a value of lattice constant as 'a'. The nearest neighbour distance in the material is:
1. $a$
2. $\sqrt{3}\frac{a}{2}$
3. $\frac{a}{\sqrt{2}}$
4. $\frac{a}{2}$

In FCC-lattice nearest neighbour distance is $\frac{a}{\sqrt{2}}$

Hence, answer is (C)

5. Light of wavelength 1.5 $\mu m$ incident on a material with a characteristic Raman frequency of $20\times 10^{12}$ Hz results in a Stokes shifted line of wavelength:
1. 1.47 $\mu m$
2. 1.57 $\mu m$
3. 1.67 $\mu m$
4. 1.77 $\mu m$

Raman shift in terms of wavelength is given by $$\frac{1}{\Delta\lambda}=\frac{1}{\lambda_0}-\frac{1}{\lambda_1}$$ $$\Delta\lambda=\frac{\lambda_0\lambda_1}{\lambda_1-\lambda_0}$$ where, $\lambda_0$ is wavelength of incident light, $\lambda_1$ is the Raman spectrum wavelength. Here, $\lambda_0=1.5\mu m$ $\lambda_1=\frac{c}{\nu}=\frac{3\times10^8}{20\times 10^{12}}=15\mu m$ $$\Delta\lambda=\frac{1.5\times15}{15-1.5}=1.67\mu m$$

Hence, answer is (C)