Quantum Mechanics MCQ Quiz - Objective Question with Answer for Quantum Mechanics - Download Free PDF

Calculation:

ΔE_ee ∝ (g_eg_e) / (m_e²a³) = g_e² / (m_e²a³)

ΔE_H ∝ (g_pg_e) / (m_pm_ea³)

⇒ ΔE_ee / ΔE_H = (g_e² / m_e²) ÷ (g_pg_e / m_pm_e) = (g_e / g_p) × (m_p / m_e)

Given:

g_p = 5.59

g_e ≈ 2

m_p / m_e ≈ 1836

So: ΔE_ee / ΔE_H = (2 / 5.59) × 1836 ≈ 656.6

ΔE_ee / ΔE_H = (2 / 5.59) × 1836 ≈ 656.6

Thus, energy increases by a factor of ≈ 656.6, and since energy and wavelength are inversely related (ΔE = hc / λ):

λ_ee / λ_H = ΔE_H / ΔE_ee ≈ 1 / 656.6

λ_ee = 21 cm / 656.6 ≈ 0.032 cm = 3.2 mm

Calculation:

The energy E of the particle is given by:

E = (ℎ² k²) / (2m) = (ℎ² π²) / (2m r₀²)

To find the pressure, we use the thermodynamic relation: P = − (∂E / ∂V), where V is the volume of the sphere.

The volume V of the sphere is:
V = (4/3) π r₀³

dE / dr₀ = ∂/∂r₀ [ (ℎ² π²) / (2m r₀²) ] = − (ℎ² π²) / (m r₀³)

dV / dr₀ = ∂/∂r₀ [ (4/3) π r₀³ ] = 4 π r₀²

⇒ dE / dV = (dE / dr₀) / (dV / dr₀) = [ − (ℎ² π²) / (m r₀³) ] / (4 π r₀²) = − (ℎ² π) / (4m r₀⁵)

P = − dE / dV = (ℎ² π) / (4m r₀⁵)

Thus, the pressure exerted on the cavity wall is given by option 1.

Calculation:

 = (3 / a²) ( S₁ · a )( S₂ · a ) − S₁ · S₂

⇒ Â = (3 / a²) a² ( S₁ · n̂ )( S₂ · n̂ ) − (1/2)(S² − S₁² − S₂²) = 3m₁m₂ − (1/2)(S(S+1) − S₁(S₁+1) − S₂(S₂+1))

⇒ S₁ = 1/2, S₂ = 1/2, S = 0, 1 and m₁ = 1/2, −1/2; m₂ = 1/2, −1/2

For S = 0: ⇒ Â = 3m₁m₂ − (1/2)(0 − 3/2) ⇒ = 3m₁m₂ + 3/4 ⇒ Hence S = 0 is singlet state so

• For m₁ = 1/2, m₂ = −1/2 or m₁ = −1/2, m₂ = 1/2 ⇒ λ₁ = −3/4 − (1/2)(0 − 3/2) = 0

For S = 1: ⇒ 3m₁m₂ − (1/2)(2 − 3/2) = 3m₁m₂ − 1/4

• m₁ = 1/2, m₂ = 1/2 ⇒ λ₂ = (3/4) − 1/4 = 1/2
• m₁ = −1/2, m₂ = −1/2 ⇒ λ₃ = (3/4) − 1/4 = 1/2
• m₁ = 1/2, m₂ = −1/2 and m₁ = −1/2, m₂ = 1/2 ⇒ λ₄ = (−3/4) − 1/4 = −1

Solution:

ψ_L = A sin(kx), x < a   and   ψ_R = B sin(k(x − 2a))

At x = a, wave function is continuous ⇒ A sin(ka) = −B sin(ka)

⇒ A = −B

⇒ − (ℎ² / 2m) (d²ψ / dx²) + V₀ δ(x − a) ψ = Eψ       integrating both sides

⇒ − (ℎ² / 2m) ∫_a−ε^a+ε d²ψ/dx² dx + V₀ ∫_a−ε^a+ε δ(x − a) ψ dx = ∫_a−ε^a+ε Eψ dx

⇒ (− ℎ² / 2m) (dψ/dx)_a+ε^a−ε + V₀ ψ(a) = 0

⇒ (− ℎ² / 2m) (−Ak cos(ka) − kA cos(ka)) + V₀ sin(ka) = 0

⇒ (ℎ² / 2m)(2kA cos(ka)) + V₀ sin(ka) = 0

⇒ (ℎ² / m) k cos(ka) = − V₀ sin(ka)   ⇒   tan(ka) = (−ℎ²k) / (mV₀)   where   k = √(2mE / ℎ²)

⇒ tan( a √(2mE) / ℎ ) = − (ℎ / V₀) √(2E / m)

Solution:

ψ = √γ exp(−γ|x|), −∞ < x < ∞ where γ = √(−2mE / ℏ²)

Energy E = −(mV₀²) / (2ℏ²)    so γ = √(−2mV₀² / ℏ²) = mV₀ / ℏ²

Probability to find x < |a|² = ∫_−a^a |ψ|² dx = (1/4) = γ ∫_−a^a exp(−2γx) dx = 2γ ∫₀^a exp(−2γx) dx = 1/4

⇒ (2mV₀ / ℏ²) × (e^−2γa − 1) / (−2γ) = −exp(−2γa) + 1 = 1/4

⇒ −exp(−2γa) = −3/4 ⇒ exp(−2γa) = 3/4 ⇒ ln(3/4) = −2γa

⇒ a = −(1 / 2γ) ln(3/4) = (1 / 2) × (ℏ² / mV₀) ln(4/3)

Final Answer: a = (ℏ² / 2mV₀) ln(4/3)

Concept:

We are using properties of angular momentum viz-\(=^2+^2+^2\) and by putting given values of \(L\) and \(L_z\) we can find value of \(L_x\).

Explanation:

 Given,

•  \(L^2|\phi>=6\hbar^2|\phi>\)
• \(L_z|\phi>=2\hbar|\phi>\)

Here L is angular momentum and \(\hbar\) is Planck's constant

Using angular momentum formula, we can write expectation values of angular momentum as

• \(=^2+^2+^2\)

Applying ket, bra operator on angular momentum operator, we get

• \(<\phi|L^2|\phi>=<\phi|L_x^2|\phi>+<\phi|L_y^2|\phi>+<\phi|L_z^2|\phi>\)

• \(<\phi|L^2|\phi>=<\phi|L_x^2|\phi>+<\phi|L_y^2|\phi>+<\phi|L_z^2|\phi>\)

Using, \(<\phi|\phi>=1\), and putting given values of \(L\) and\(L_z\) we get,

• \(6\hbar^2=++4\hbar^2\)
• Now, \(\)\(=\)

we get, \(\frac {6\hbar^2-4\hbar^2} {2}=\)

• \(=\hbar^2\)

The correct answer is \(=\hbar^2\)

Concept:

The given Hamiltonian is for the Anharmonic Oscillator. We will compare the given Hamiltonian with the equation of the Hamiltonian Anharmonic oscillator.

•  \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+2 m ω^2 y^2\)
• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+\frac {1}{2}m (2ω)^2 y^2\)

Explanation:

• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+2 m ω^2 y^2\)
• \(H=\frac{p_x^2}{2 m}+\frac{p_y^2}{2 m}+\frac{1}{2} m ω^2 x^2+\frac {1}{2}m (2ω)^2 y^2\)
• \(\omega_x=\omega\), \(\omega_y=2\omega\)
• \(E= (n_x+\frac{1}{2})\hbar\omega_x+(n_y+\frac{1}{2})\hbar\omega_y\)

This is the formula for energy in an oscillator in two-dimension.

Now, \(E=\)\(\frac{27}{2} \hbar \omega\)(given)

• \(\frac{27}{2} \hbar \omega\)\(= (n_x+\frac{1}{2})\hbar\omega_x+(n_y+\frac{1}{2})\hbar\omega_y\)
• Substitute values of \(\omega_x\) and \(\omega_y\) in terms of \(\omega\)
• \(\frac{27}{2} \hbar \omega\)\(= (n_x+\frac{1}{2})\hbar\omega+2(n_y+\frac{1}{2})\hbar\omega\)
• \(\frac {27}{2}=n_x+\frac{1} {2}+2n_y+1\)
• \(12=n_x+2n_y\)
• \(n_x\) should be even to satisfy the equation.
• \(n_x\)\(=0,2,4,6,8,10,12\) ; \(n_y=6,5,4,3,2,1,0\)
• \((n_x,n_y)=(0,6),(2,5),(4,4), (6,3), (8,2),(10,1), (12,0)\)
• So, the degeneracy of the energy level \(E=\)\(\frac{27}{2} \hbar \omega\) is 7.

So, the correct answer is 7.

Explanation:

• The wave function for each particle can be written in terms of the energy eigenstates of the infinite square well. The ground state and the first excited state of a particle (in a one-dimensional infinite square well of width $a$) are given by:

\(\psi_0(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right)\) , for the ground state, and

\(\psi_1(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{2\pi x}{a}\right)\) , for the first excited state.

• The expectation value of the product of the position operators \(x_1\) and \(x_2\) can be calculated with the help of the integral:

\(\langle x_1 x_2 \rangle = \int dx_1 \int dx_2\) \(\psi_0(x_1)\psi_1(x_2) x_1 x_2 \psi_0(x_1) \psi_1(x_2)\),

• Where the integration is over the range [0, a] for both \(x_1\) and \(x_2\).
• However, since the particles are distinguishable and non-interacting, we can write \(\langle x_1 x_2 \rangle = \langle x_1 \rangle \langle x_2 \rangle \), which simplifies the integrations to:

\(\langle x_1 \rangle = \int dx_1 \psi_0(x_1) x_1 \psi_0(x_1)\) , and \(\langle x_2 \rangle = \int dx_2 \psi_1(x_2) x_2 \psi_1(x_2)\).

• Calculation of these integrals gives: \(\langle x_1 \rangle = \frac{a}{2}\), and \(\langle x_2 \rangle = \frac{a}{2}\).
• So, the expectation value of the product of \(x_1\) and \( x_2\) is just the product of the expectation values of \(x_1\) and \(x_2\): \(\langle x_1 x_2 \rangle = \langle x_1 \rangle \langle x_2 \rangle = \frac{a^2}{4}\)

Solution-Option-1

Concept:

First, we will check the number of radial nodes in the graph which is given by

• Number of radial nodes \(=n-l-1\)
• Then secondly we will check the value of probability of electron at \(r=2a\).

Calculation-

• Given- \(n=2\) and \(l=0\)
• \(R_{20}=N\left(1-\frac{r}{2 a}\right) e^{-\frac{r}{2 a}}\)
• Radial nodes\(=n-l-1\) \(=2-0-1=1\)
• So, there will be only one node in the graph. Either option 1 is correct or option 2 is correct.
• Now we will check the probability of finding an electron at \(r=2a\).
• At, \(R_{20}=N(1-\frac{r}{2a})\mathrm{e}^\frac{-r}{2a}\)   
• \(P_r=|R_{20}|^2=N^2(1-\frac{r}{2a})^2\mathrm{e}^{\frac{-r}{a}}\)
• At \(r=2a\), \(P_r=0\)
• The probability of finding an electron at r=2a" id="MathJax-Element-14-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a" id="MathJax-Element-270-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a r=2a $r=2a$  is zero.
• So, graph 1 satisfies this condition.

So, the correct answer is Graph-1.

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 \(\frac{1}{4} N^2 E_0\).

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).

Explanation:

Let us reconsider the system of equations:

\(\frac{dA}{dt} = -iωB\) and \(\frac{dB}{dt} = iωA.\)

By differentiating the first equation again, \( \frac{d²A}{dt²} = -iω \frac{dB}{dt}.\)

• Substituting the second equation into this results in \(\frac{d²A}{dt²} = -ω² A.\)
• This differential equation is a simple harmonic one, but with a key difference: there is no negative in front of ω², leading to hyperbolic solutions.
• Specifically, we find A(t) = Csinh(ωt), for some constant C.
• Given that the expectation value \(〈A⟩_ψ(t) = 0\) at t = 0, we find \(C = 0.\)
• Thus, in general, B(t) has to be in the form of cosh(ωt), to meet the commutation relations. Finally, given that \(〈B⟩_ψ (0) = i,\) we need to multiply cosh(ωt) by i.
• So the time-evolved expectation value is \( 〈A⟩_ψ(t) = 〈ψ|A(t)| ψ⟩ = 〈ψ|Csinh(ωt)| ψ⟩ = sinh(ωt)\)

Concept: 

 Energy in first perturbation is given by

• \(E_1^{(1)}=<\psi|H^\prime|\psi>\)
• We know that, for an infinite potential well, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) 

Explanation:

Given-

• \(H^{\prime}=ϵ \cos \left(\frac{\pi x}{L}\right)\) limits from \(\frac {-L} {2}\) to \(\frac {+L} {2}\)

We know that, for an infinite potential well, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) 

• \(E_1^{(1)}=<\psi|H^\prime|\psi>\)
• \(E_1^{(1)}=\int\limits_\frac{-L} {2}^\frac{+L}{2} |\psi|^2 H^{\prime}dx\)

Now, \(\psi=\sqrt{\frac {2} {L}}cos\frac {\pi x} {L}\) and \(H^{\prime}=ϵ \cos \left(\frac{\pi x}{L}\right)\), put these values in first order energy perturbation equation, we get,

• \(E_1^{(1)}=\int\limits_\frac{-L} {2}^\frac{+L}{2} [\sqrt {\frac{2} {L}}cos\frac{\pi x}{L}]^2 .\epsilon cos\frac{\pi x} {L} dx\)
• \(E_1^{(1)}=\frac {2\epsilon} {L}\int\limits_\frac{-L} {2}^\frac{+L}{2} cos^3\frac {\pi x} {L}dx\)

Now for changing the limit from(\(\frac{-L} {2}\) to \(\frac {+L} {2}\)) to (\(0\) to \(\frac {+L} {2}\))

•  \(E_1^{(1)}=\frac {2\epsilon} {L}\times 2\int\limits_0^\frac {+L}{2}\cos^3\frac {\pi x} {L}dx\)

Using the trignometric formula for \(cos^3\frac {\pi x} {L}\),

• \(cos^3\frac{\pi x }{L}=\frac{[cos(\frac{3\pi x}{L})+3cos(\frac {\pi x}{L})]} {4}\)
• Substituting this value, we get,
• \(E_1^{(1)}=\frac {4\epsilon} {L}\int\limits_0^\frac {+L}{2}\frac{[cos(\frac{3\pi x}{L})+3cos(\frac {\pi x}{L})]} {4}\)
• \(E_1^{(1)}=\frac {\epsilon} {L} [(\frac {Sin\frac {3\pi x}{L}} {3\pi/L})|_0^{L/2} + 3(\frac {Sin\frac{\pi x}{L}} {\pi/L})|_0^{L/2}\)

• \(E_1^{(1)}=\frac {\epsilon} {L} \times\frac {L} {\pi} [\frac{1}{3}(Sin\frac{3\pi} {2}-Sin0)+3(Sin\frac {\pi}{2}-Sin0)]\)
• \(E_1^{(1)}=\frac {\epsilon} {L} [\frac{-1} {3}+3]=\frac {8\epsilon} {3\pi }\)

So, the correct answer is \(\frac{8 \epsilon}{3 \pi}\).

Concept:

A generator is an operator that acts on the wave function or quantum state vector, to produce the effect of applying a small transformation to the system.

Calculation:

q → q' = (1 + ϵ)q

p → p' = (1 - ϵ)p

If G is the generator then 

p' - p = δ p_j 

= - ϵ \({\partial G \over \partial q_j}\)

= - ϵ p

q' - q = δ q_i 

= ϵ \({\partial G \over \partial p_j}\)

= ϵ p

Now G = qp

- ϵ \({\partial G\over \partial q} \) = - ϵ p = δ p

ϵ \({\partial G\over \partial p} \) = - ϵ q = δ q

The correct answer is option (2).