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

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:

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).

CONCEPT: 

1. Energy Operator:

The energy (or Hamiltonian) operator in one dimension for a particle in a box is given by:

\(\hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}\)

2. Expectation Value of Energy:

The expectation value of energy 〈 E 〉  is given by:

〈 E 〉 = \(\int_0^1 ψ^ *(x) \hat{H} ψ(x) \, dx\)

Calculation -

1. Wavefunction:
   
\(ψ(x) = \sqrt{\frac{8}{5}} \sin(\pi x)(1 + \cos(\pi x))\)

2. Second Derivative:
   
\(\frac{d}{dx} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = \pi \cos(\pi x)(1 + \cos(\pi x)) - \pi \sin^2(\pi x)\)
   
\(\frac{d^2}{dx^2} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = -\pi^2 \sin(\pi x)(1 + \cos(\pi x)) - 2\pi^2 \cos(\pi x) \sin(\pi x) \)

Simplifying:

\(\frac{d^2}{dx^2} \left( \sin(\pi x)(1 + \cos(\pi x)) \right) = -\pi^2 \sin(\pi x) (1 + 2\cos(\pi x) + \cos^2(\pi x)) \)

3. Hamiltonian Acting on ψ(x):
   
\(\hat{H} ψ(x) = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} ψ(x) \)   

4. Expectation Value Integral:
   
〈 E 〉 = \(\int_0^1 ψ(x) \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} ψ(x)\right) dx\)

The normalized wavefunction ψ(x) given is a linear combination of eigenfunctions of the infinite potential well.

For an infinite potential well, the eigenfunctions are

\(ψ_n(x) = \sqrt{2} \sin(n \pi x)\) with energy eigenvalues \(E_n = \frac{n^2 \pi^2 \hbar^2}{2m} .\)

Comparing the given wavefunction:

\(ψ(x) = \sqrt{\frac{8}{5}} \sin(\pi x)(1 + \cos(\pi x))\)

This is equivalent to a superposition of the first and second eigenstates.

The coefficients and normalization ensure that this wavefunction is a proper eigenstate mixture.

Energy Calculation:

By symmetry and orthogonality of the eigenfunctions, the expectation value of energy 〈 E 〉 is the weighted sum of eigenvalues:

〈 E 〉 = \(a_1^2 E_1 + a_2^2 E_2\)

Given:

\(E_1 = \frac{\pi^2 \hbar^2}{2m}, \quad E_2 = \frac{4 \pi^2 \hbar^2}{2m}\)

The weights a₁ , a₂ are found from normalization. Simplifying using known integrals, we obtain the correct weighted sum.

Finally, the result for this particular problem (via solving) yields the expectation value \(\frac{4\pi^2 \hbar^2}{5m}\)

Therefore, the correct answer is (2).

CONCEPT:

For an infinite potential well (as shown)

The wave inside Ψ_in (≠ 0)  is oscillatory and the wave outside Ψ_out = 0

The energy of the n^th state is given by, 

\(⇒ E_n = n^2 \left( \frac{\pi^2 \hbar^2}{2mL^2} \right) \)

For a finite potential well (as shown)

The wave inside Ψin (≠ 0)  is oscillatory and the wave outside Ψout is damped (exponentially decreasing).

So, the energy will be more and more lost due to penetration in the walls. 

So, the energy in the finite potential well in less than that of infinite potential well. 

EXPLANATION:

From the knowledge above the possible waveforms for the potential wells A, B, and C are shown below

Case A:

The wave inside Ψin (≠ 0)  is oscillatory and the wave outside Ψout is damped (exponentially decreasing).

As the height of the wall in LHS is more than RHS so the penetration in the RHS will be deeper. 

So, the energy is lost due to penetration in both sides of the waves. 

Case B: 

The wave inside Ψin (≠ 0)  is oscillatory and the wave on LHS Ψout  = 0 and on the RHS is Ψout  = damped (exponentially decreasing).

As the height of the wall in LHS is infinity and the RHS is V₀ 

In LHS there will be no penetration. 

So, the energy is lost due to penetration in one sides on the wall. 

So, energy is more compared to case A.

Case C: 

The wave inside Ψin (≠ 0)  is oscillatory and the wave outside Ψout = 0

Here, no penetration outside so the least energy is lost. so most energy of the states among all. 

So, the correct order of energies is EC > EB > EA.

Hence the correct answer is option 1.