Quantum Mechanics MCQ Quiz in తెలుగు - Objective Question with Answer for Quantum Mechanics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

 $E_n=n{E}_0$(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 $E_i=2\times 0+2\times E_0+2\times 2E_0+2\times 3E_0+-------+2\times (\frac{N-2}{2})E_0$
• $E_i=2E_0[1+2+3+-----------+\frac{N-2}{2}]$
• 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^2{E}_0}{4}-\frac{N{E}_0}{2}$

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

• $E_f=1\times 0+1\times E_0+1\times 2E_0+1\times 3E_0+...+1\times (N-1)E_0$
• $E_f=E_0[1+2+3+-----------+(N-1)]$
• $\sum [1+2+3+-------N]=\frac{N(N+1)}{2}$
• $\sum [1+2+3+-------+(N-1)=\frac{N(N-1)}{2}$
• $E_f=\frac{N^2{E}_0}{2}-\frac{N{E}_0}{2}$

The change in ground state energy is $E_f-E_i=\frac{N^2{E}_0}{2}-\frac{N{E}_0}{2}-\frac{N^2{E}_0}{4}+\frac{N{E}_0}{2}=\frac{N^2{E}_0}{4}$

So, the correct answer is $\frac{1}{4}{N}^2{E}_0$.

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

$\psi (x,t=0)=|{\phi }_1⟩-|{\phi }_3⟩$

$\psi (x,t)=|{\phi }_1⟩e^{\frac{-i{E}_{t}t}{\hslash }}$ - $|{\phi }_3⟩e^{\frac{-i{E}_{3}t}{\hslash }}$

$E_1=\frac{{\pi }^2{\hslash }^2}{2m{L}^2}{E}_3$ = $\frac{9{\pi }^2{\hslash }^2}{2m{L}^2}t=\frac{m{L}^2}{4\pi \hslash }$

$\psi (x,t)=|{\phi }_1⟩e^{\frac{-i\pi }{8}}-|{\phi }_3⟩e^{\frac{-9i\pi }{8}}$ = $e^{\frac{-i\pi }{8}}(|{\phi }_1⟩-|{\phi }_3⟩e^{-i\pi })$

$=e^{\frac{-i\pi }{8}}(|{\phi }_1⟩+|{\phi }_3⟩)=e^{\frac{-i\pi }{8}}$$(\sqrt{\frac{2}{L}}\sin \frac{\pi x}{L}+\sqrt{\frac{2}{L}}\sin \frac{3\pi x}{L})$

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 $L^2\psi =l(l+1){\hslash }^2\psi$, 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:

$E_{n_x,n_y}=\frac{(n_x^2+n_y^2){\pi }^2{\hslash }^2}{2m{l}^2}$. 

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:

$E=2\times \frac{2{\pi }^2{\hslash }^2}{2m{l}^2}+2\times \frac{5{\pi }^2{\hslash }^2}{2m{l}^2}+2\times \frac{5{\pi }^2{\hslash }^2}{2m{l}^2}+1\times \frac{8{\pi }^2{\hslash }^2}{2m{l}^2}$.

Simplifying it a bit we get:

$E=\frac{32{\pi }^2{\hslash }^2}{2m{l}^2}$.

Hence the correct option is option1).

Explanation:

 We have been given with ${\delta }_0={90}^0=\frac{\pi }{2}$and wavevector is $\sqrt{3\pi }f{m}^{-1}$. 

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:

$\sigma ==\frac{4\pi }{3\pi }f{m}^2=1.33f{m}^2$.

The correct option is option 3).

Explanation:

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

$[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k,[{\sigma }_i,{\sigma }_j{]}_a=2I{\delta }_{ij}$ , where a subscript is for anti-commutation.

So, using the above relation we get:

${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ .

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:

$3{\sigma }_x{\sigma }_y+5{\sigma }_y{\sigma }_x=3i{\sigma }_z-5i{\sigma }_z=-2i{\sigma }_z$ .

The correct option is option 2).

Explanation:

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

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

Therefore $S^2=S_1^2+S_2^2+2{\overrightarrow{S}}_1\cdot {\overrightarrow{S}}_2$.

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:

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

The correct option is option 3).

Explanation:

At t=0 we have $|\varphi >=\frac{1}{\sqrt{2}}(|{\psi }_{+}>+|{\psi }_{-}>)$. 

Now at some other time we will have the state evolved as $|{\varphi }_1>=\frac{1}{\sqrt{2}}(|{\psi }_{+}>e^{-\frac{iϵt}{\hslash }}+|{\psi }_{-}>e^{\frac{iϵt}{\hslash }})$.

If $t=h/6ϵ$ we have $|{\varphi }_1>=\frac{1}{\sqrt{2}}(|{\psi }_{+}>e^{-\frac{i\pi }{3}}+|{\psi }_{-}>e^{\frac{i\pi }{3}})$.

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

$|<{\varphi }_1|\varphi >{|}^2=|\frac{e^{-i\pi /3}+e^{i\pi /3}}{2}{|}^2=|\cos (\frac{\pi }{3}){|}^2=\frac{1}{4}=0.25$.

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 $\alpha \to 0$ , we have columbic potential.

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

which implies that $f(\theta )\propto cose{c}^4\frac{\theta }{4}$ . .

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

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

The correct option is option 2).

Concept:

The momentum operator is given by

p = - ih $\frac{\partial }{\partial x}$

where h is the Plank constant.

Calculation:

e^{$\frac{iaP}{h}$} |x>

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

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

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

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

The correct answer is an option (3).