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

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పొందండి Quantum Mechanics సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Quantum Mechanics MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Quantum Mechanics MCQ Objective Questions

Top Quantum Mechanics MCQ Objective Questions

Quantum Mechanics Question 1:

At t = 0 , the wavefunction of an otherwise free particle confined between two infinite walls at x = 0 and x = L is ψ(x, t = 0) = \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}-\sin \frac{3 \pi x}{L}\right)\). Its wave function at a later time \(t=\frac{m L^2}{4 \pi h}\) is

  1. \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}-\sin \frac{3 \pi x}{L}\right) e^{i \pi / 6}\)
  2. \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}+\sin \frac{3 \pi x}{L}\right) e^{-i \pi / 6}\)       
  3. \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}-\sin \frac{3 \pi x}{L}\right) e^{-i \pi / 8}\)  
  4. \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}+\sin \frac{3 \pi x}{L}\right) e^{-i \pi / 8}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt{\frac{2}{L}}\left(\sin \frac{\pi x}{L}+\sin \frac{3 \pi x}{L}\right) e^{-i \pi / 8}\)

Quantum Mechanics Question 1 Detailed Solution

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

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

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

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

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

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

Quantum Mechanics Question 2:

A hydrogen atom is in the state \(\psi = \sqrt{\frac{8}{21}} \psi_{200} -\sqrt{\frac{3}{7}} \psi_{310} + \sqrt{\frac{4}{21}} \psi_{321} \), where n, l, m in \(\psi_{nlm} \) denote the principle orbit and magnetic quantum numbers, respectively. What is the expectation value of L2 in units of \(\frac{\hbar^2}{2} \) ?

  1. 2
  2. 4
  3. 8
  4. 1

Answer (Detailed Solution Below)

Option 2 : 4

Quantum Mechanics Question 2 Detailed Solution

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).

Quantum Mechanics Question 3:

A two-dimensional square rigid box of side l contains 7 non-interacting electrons at T= 0K. The mass of the electron is m. The ground state energy of the system of electrons in units of \(\frac{\pi^2 \hbar^2}{2 m l^2} \) is?

  1. 32
  2. 24
  3. 14
  4. 7

Answer (Detailed Solution Below)

Option 1 : 32

Quantum Mechanics Question 3 Detailed Solution

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).

Quantum Mechanics Question 4:

Consider an elastic scattering of particles in l=0 states. If the corresponding phase shift is \(\delta_0 \) is 900and magnitude of the incident wave vector is equal to \(\sqrt{3 \pi} fm^{-1} \), then the total scattering cross section in units of fm-2 is? (Correct up to 2 decimal places)

  1. 1.32
  2. 1.34
  3. 1.33
  4. 1.35

Answer (Detailed Solution Below)

Option 3 : 1.33

Quantum Mechanics Question 4 Detailed Solution

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).

Quantum Mechanics Question 5:

If \(\vec{S}_1 \) and \(\vec{S}_2 \) are the spin operators of the two electrons of a He atom, the value of \(<\vec{S}_1 \cdot \vec{S}_2 > \) for the ground state is ?

  1. \(\frac{\hbar^2}{4} \)
  2. \(-\frac{3}{2} \hbar^2 \)
  3. \(-\frac{3}{4} \hbar^2 \)
  4. 0

Answer (Detailed Solution Below)

Option 3 : \(-\frac{3}{4} \hbar^2 \)

Quantum Mechanics Question 5 Detailed Solution

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).

Quantum Mechanics Question 6:

A two-state quantum system has energy eigenvalues is \(\pm \epsilon \) corresponding to the normalized states \(|\psi_{\pm} > \). At time t=0, the system is in quantum state \(\frac{1}{\sqrt{2}} (|\psi_{+}> + |\psi_{-}> ) \). The probability that the system will be in the same state at \(t= h/ 6\epsilon \) is

  1. 0.2
  2. 0.25
  3. 0.3
  4. 0.5

Answer (Detailed Solution Below)

Option 2 : 0.25

Quantum Mechanics Question 6 Detailed Solution

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).

Quantum Mechanics Question 7:

Consider potential \(U (r)= - U_0 \frac{e^{-\alpha r}}{r} \). Both \(\alpha \) and \(U_0 \) are constants. According to first Born approximation, the elastic scattering amplitude calculated with \(U(r) \) for momentum transfer q and \(\alpha \rightarrow 0 \) is proportional to?

  1. \(q \)
  2. \(q^{-2} \)
  3. \(q^{-1} \)
  4. \(q^{2} \)

Answer (Detailed Solution Below)

Option 2 : \(q^{-2} \)

Quantum Mechanics Question 7 Detailed Solution

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).

Quantum Mechanics Question 8:

Let \(\hat x\) and \({\rm{\hat p}}\) denote position and momentum operators obeying the commutation relation \(\left[ {{\rm{\hat x,}}\,{\rm{\hat p}}} \right]\) = ih. If |x denotes an eigenstate of \({{\rm{\hat x}}}\) corresponding to the eigenvalue x, then \({{\rm{e}}^{{\rm{ia\hat p/h}}}}\left| x \right\rangle \) is

  1. an eigenstate of \(\hat x\) corresponding to the eigenvalue x
  2. an eigenstate of \(\hat x\) corresponding to the eigenvalue (x + a)
  3. an eigenstate of \(\hat x\) corresponding to the eigenvalue (x − a)
  4. not an eigenstate of \(\hat x\)

Answer (Detailed Solution Below)

Option 3 : an eigenstate of \(\hat x\) corresponding to the eigenvalue (x − a)

Quantum Mechanics Question 8 Detailed Solution

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).

Quantum Mechanics Question 9:

A particle of mass m is confined to a box of unit length in one dimension. It is described by the wavefunction ψ(x) = \(\sqrt {\frac{8}{5}} \) sin πx (1 + соs πx) for 0 ≤ x ≤ 1, and zero outside this interval. The expectation value of energy in this state is

  1. \(\frac{{4{{\rm{\pi }}^2}}}{{3{\rm{m}}}}\) \(\hbar^2\)
  2. \(\frac{{4{{\rm{\pi }}^2}}}{{5{\rm{m}}}}\) \(\hbar^2\)
  3. \(\frac{{2{{\rm{\pi }}^2}}}{{4{\rm{m}}}}\) \(\hbar^2\)
  4. \(\frac{{8{{\rm{\pi }}^2}}}{{5{\rm{m}}}}\) \(\hbar^2\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{{4{{\rm{\pi }}^2}}}{{5{\rm{m}}}}\) \(\hbar^2\)

Quantum Mechanics Question 9 Detailed Solution

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 a, 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).

Quantum Mechanics Question 10:

For the one dimensional potential wells A, B and C, as shown in the figure, let EA, EB and EC denote the ground state energies of a particle, respectively.

F2 Technical Mrunal 24.02.2023 D1

The correct ordering of the energies is

  1. EC > EB > EA
  2. EA > EB > EC
  3. EB > EC > EA
  4. EB > EA > EC

Answer (Detailed Solution Below)

Option 1 : EC > EB > EA

Quantum Mechanics Question 10 Detailed Solution

CONCEPT:

For an infinite potential well (as shown)

F2 Technical Mrunal 24.02.2023 D2

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

The energy of the nth state is given by, 

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

For a finite potential well (as shown)

F2 Technical Mrunal 24.02.2023 D3

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:

F2 Technical Mrunal 24.02.2023 D4

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.

F2 Technical Mrunal 24.02.2023 D5

Case C: 

F2 Technical Mrunal 24.02.2023 D6

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.

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