Quantum Mechanics MCQ Quiz in বাংলা - Objective Question with Answer for Quantum Mechanics - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

 \(E_n=nE_0\)(given)

Case-1-Initial ground state energy without polarization

According to Pauli Exclusion principle, only two electrons filled in one state.

• Initial ground state energy \(E_i=2\times0+2\times E_0+2\times 2E_0+2\times3E_0+-------+2\times (\frac{N-2}{2})E_0\)
• \(E_i=2E_0[1+2+3+-----------+\frac {N-2}{2}]\)
• Now, \(\sum[1+2+3+-------N]=\frac{N(N+1)}{2}\)
• \(\sum[1+2+3+-------\frac{N-2} {2}]=\frac {(\frac{N-2}{2}) (\frac {N} {2})} {2}\)
• \(E_i=2E_0\times \)\(\frac {(\frac{N-2}{2}) (\frac {N} {2})} {2}\)\(=\frac{N^2E_0}{4}-\frac{NE_0} {2}\)

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

• \(E_f=1\times0+1\times E_0+1\times 2E_0+1\times3E_0+...+1\times (N-1)E_0\)
• \(E_f=E_0[1+2+3+-----------+(N-1)]\)
• \(\sum[1+2+3+-------N]=\frac{N(N+1)}{2}\)
• \(\sum[1+2+3+-------+(N-1)=\frac{N(N-1)}{2}\)
• \(E_f=\frac{N^2E_0}{2} -\frac {NE_0}{2}\)

The change in ground state energy is \(E_f-E_i=\frac{N^2 E_0} {2}-\frac {NE_0}{2}-\frac {N^2E_0} {4}+\frac {NE_0}{2}=\frac {N^2 E_0}{4}\)

So, the correct answer is \(\frac{1}{4} N^2 E_0\).

ψ(x, t = 0) = \(\left(\sqrt{\frac{2}{L}} \sin \frac{\pi x}{L}-\sqrt{\frac{2}{L}} \sin \frac{3 \pi x}{L}\right)\)

\(\psi(x, t=0)=\left|\varphi_1\right\rangle-\left|\varphi_3\right\rangle\)

\(\psi(x, t)=\left|\varphi_1\right\rangle e^{\frac{-i E_t t}{\hbar}}\) - \(\left|\varphi_3\right\rangle e^{\frac{-i E_3 t}{\hbar}} \)

\(E_1=\frac{\pi^2 \hbar^2}{2 m L^2} E_3\) = \(\frac{9 \pi^2 \hbar^2}{2 m L^2} t=\frac{m L^2}{4 \pi \hbar}\)

\(\psi(x, t)=\left|\varphi_1\right\rangle e^{\frac{-i \pi}{8}}-\left|\varphi_3\right\rangle e^{\frac{-9 i \pi}{8}}\) = \(e^{\frac{-i \pi}{8}}\left(\left|\varphi_1\right\rangle-\left|\varphi_3\right\rangle e^{-i \pi}\right)\)

\(=e^{\frac{-i \pi}{8}}\left(\left|\varphi_1\right\rangle+\left|\varphi_3\right\rangle\right)=e^{\frac{-i \pi}{8}}\)\(\left(\sqrt{\frac{2}{L}} \sin \frac{\pi x}{L}+\sqrt{\frac{2}{L}} \sin \frac{3 \pi x}{L}\right)\)

Explanation:

We have \(\psi = \sqrt{\frac{8}{21}} \psi_{200} -\sqrt{\frac{3}{7}} \psi_{310} + \sqrt{\frac{4}{21}} \psi_{321} \)

Now <\(\psi| \psi \)> =1. Hence normalized.

Now we have to find the < L²> = \(\frac{8}{21} < \psi_{200}|L^2|\psi_{200}> +\frac{3}{7} < \psi_{310}|L^2|\psi_{310}>+ \frac{4}{21} < \psi_{321}|L^2|\psi_{321}> \) .

Now using the relation that \(L^2 \psi = l (l+1) \hbar^2 \psi \), we get:

\(=0+\frac{6}{7} \hbar^2 + \frac{8}{7} \hbar^2 = 2 \hbar^2 = 4\times\frac{\hbar^2}{2}\).

So the correct option is option 2).

Explanation:

WE are given a 2-dimensional box of length l.

There are electrons in it of mass m.

We have to find the ground state energy of 7 electrons.

So, we know that in 2-dimensional box energy eigen value is given by:

\(E_{n_x, n_y}= \frac{(n_x^2 + n_y^2) \pi^2 \hbar^2}{2 ml^2} \). 

For the ground state energy of 7 electrons the arrangement can be:

\(E= 2 \times E_{1,1} + 2 \times E_{2,1} + 2\times E_{1,2} + 1 \times E_{2,2} \) .

hence the energy would be:

\(E= 2 \times \frac{2\pi^2 \hbar^2}{2 ml^2} + 2 \times \frac{5\pi^2 \hbar^2}{2 ml^2} + 2 \times \frac{5 \pi^2 \hbar^2}{2 ml^2} + 1 \times \frac{ 8\pi^2 \hbar^2}{2 ml^2} \).

Simplifying it a bit we get:

\(E= \frac{32 \pi^2 \hbar^2}{2 m l^2} \).

Hence the correct option is option1).

Explanation:

 We have been given with \(\delta_0 = 90^0 = \frac{\pi}{2} \)and wavevector is \(\sqrt{3 \pi} fm^{-1} \). 

The total scattering cross-section for particle at l=0 states is given as :

\(\sigma = \frac{4 \pi }{k^2} \sin^2 \delta_0 \). Putting all the values given in the above expression we get:

\(\sigma = = \frac{4\pi}{3 \pi} fm^2 =1.33 fm^2 \).

The correct option is option 3).

Explanation:

 The commutation and anti-commutation relations of Pauli matrices are given as:

\([\sigma_i , \sigma_j] = 2 i \epsilon_{ijk} \sigma_k , [\sigma_i, \sigma_j]_a = 2 I \delta_{ij} \) , where a subscript is for anti-commutation.

So, using the above relation we get:

\(\sigma_i \sigma_j = \delta_{ij} I + i \epsilon_{ijk} \sigma_k \) .

So, using the above relation we can write:

\(\sigma_x \sigma_y = i \sigma_z , \sigma_{y} \sigma_x = - i \sigma_z \) .

Putting these in the final expression we get:

\(3 \sigma_x \sigma_y + 5 \sigma_y \sigma_x = 3 i \sigma_z -5 i \sigma_z = - 2 i \sigma_z \) .

The correct option is option 2).

Explanation:

We have two electrons hence the total spin will be: \(\vec{S} = \vec{S}_1 + \vec{S}_2 \) .

Now,  \(\vec{S}\cdot \vec{S}= (\vec{S}_1 + \vec{S}_2 ) \cdot (\vec{S}_1 + \vec{S}_2) \) .

Therefore \(S^2= S_1^2 + S_2^2 + 2 \vec{S}_1 \cdot \vec{S}_2 \).

Taking the expectation values on both the sides we get:

\( < \vec{S}_1 \cdot \vec{S}_2> = \frac{1}{2} ( -< S_1^2> -< S_2^2> ) \).

We know that \( = s_1(s_1 + 1) \hbar^2 , = s_2(s_2 + 1) \hbar^2, = s(s + 1) \hbar^2 \).

Also, we know that \(s_1 = s_2 = \frac{1}{2} \), as it is for electrons so for ground state the total spin will be s=0.

Hence putting all these in the above expression for expectation value we get:

\( < \vec{S}_1 \cdot \vec{S}_2> = \frac{\hbar^2}{2} (0 - \frac{2}{2} (\frac{1}{2} +1) ) = - \frac{3 \hbar^2}{4} \).

The correct option is option 3).

Explanation:

At t=0 we have \(|\phi> = \frac{1}{\sqrt{2}} (|\psi_{+}> + |\psi_{-}> ) \). 

Now at some other time we will have the state evolved as \(|\phi_1> = \frac{1}{\sqrt{2}} (|\psi_{+}> e^{-\frac{i\epsilon t}{\hbar}}+ |\psi_{-}> e^{\frac{i\epsilon t}{\hbar}}) \).

If \(t= h/ 6\epsilon \) we have \(|\phi_1> = \frac{1}{\sqrt{2}} (|\psi_{+}> e^{-\frac{i\pi}{3}}+ |\psi_{-}>e^{\frac{i\pi}{3}}) \).

Now in order to find the probability that this state will appear after the above stated time is;

\(|<\phi_1| \phi> |^2 =| \frac{e^{-i\pi/3} + e^{i\pi /3}}{2} |^2 = |\cos(\frac{\pi}{3})|^2 =\frac{1}{4} = 0.25 \).

We have used the property here that \(<\psi_{+}| \psi_{-}> =0 , <\psi_{-}| \psi_{+}> =0 , <\psi_{+}| \psi_{+}>=<\psi_{-}| \psi_{-}> =1 \).

The correct option is option 2).

Explanation:

For \(\alpha \rightarrow 0 \) , we have columbic potential.

For this type of potential, we have \(\sigma (\theta) \propto cosec^{4} \frac{\theta}{4} \) ,

which implies that \(f(\theta) \propto cosec^{4} \frac{\theta}{4} \) . .

Also, momentum transfer is given as \( q= 2 k \sin \frac{\theta}{2} \).

 Hence, \(f (\theta) \propto \frac{1}{q^2} \).

The correct option is option 2).

Concept:

The momentum operator is given by

p = - ih \({\partial \over \partial x}\)

where h is the Plank constant.

Calculation:

e^{\({iaP\over h}\)} |x>

= [\(\sum^{\infty}_{n=0} {1 \over n!}({iaP\over h})^n\) ]|x>

= \([\sum^{\infty}_{n=0} {1 \over n!}({iaP\over h})^n]^h \)|x>

= |x> - a∇|x> + \({1 \over 2!}\) (a∇)²|x> ... = |x-a>

X|x-a> = (x-a)|x-a>

The correct answer is an option (3).