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

The correct answer is For n = 1: P = 0.031, For n = 2: P = 0.029

EXPLANATION:

P ≈ (2Δx / L) · sin2(nπx / L).

• The formula for the probability is given as:
• Substitute the values for n = 1 and n = 2, x = 0.66L, and Δx = 0.02L:

  P = (2 × 0.02L / L) · sin2(π × 0.66)

  = 0.04 · sin2(0.66π)

  ≈ 0.031.

  P = (2 × 0.02L / L) · sin2(2π × 0.66)

  = 0.04 · sin2(1.32π)

  ≈ 0.029.

  • For n = 1:
    
  • For n = 2:
    

CONCLUSION:

The approximate probabilities are For n = 1: P = 0.031 & For n = 2: P = 0.029

CONCEPT:

Normalization of the Radial Part of a Hydrogen Atom Wave Function

• Normalization ensures that the probability of finding the electron within the entire space is equal to 1.
• The radial wave function for a hydrogen-like atom in its ground state can be written as \( R_{1,0} = N e^{-r/a_0} \), where (N) is the normalization constant, and a₀ is the Bohr radius.
• To normalize the wave function, the integral of the probability density over all space must be equal to 1:
  • \( \int_0^{\infty} |R_{1,0}|^2 r^2 \, dr = 1 \)

CALCULATION:

• The radial wave function is given by \( R_{1,0} = N e^{-r/a_0} \).
• To normalize it, we calculate:
  • \( \int_0^{\infty} |N e^{-r/a_0}|^2 r^2 \, dr = 1 \)
  • \( N^2 \int_0^{\infty} r^2 e^{-2r/a_0} \, dr = 1 \)
• Solving this integral:
  • Using the given solution, \( \int_0^{\infty} r^2 e^{-2r/a_0} \, dr = \frac{a_0^3}{8} \)
  • This gives us \(N^2 \cdot \frac{a_0^3}{8} = 1 \)
  • Solving for ( N ), we get: \( N = \frac{2}{a_0^{3/2}} \)

CONCLUSION:

• The correct normalization constant is:
  • Option (1): \( \frac{2}{a_0^{3/2}} \)

Concept:

In quantum mechanics, the expectation value of energy for an electron in a mixed state can be calculated by taking the weighted average of the energies of each state in the superposition. For the hydrogen atom, the energy of a state is given by:

Energy Formula: The energy of an electron in a hydrogen atom with principal quantum number ( n ) is:

\(E_n = -\frac{13.6 \, \text{eV}}{n^2} \).

In a mixed state, where the wave function is a combination of different states with weights, the expectation value of the energy \(〈 E 〉 \) is calculated as:

\(〈 E 〉 = \sum \left( |c_i|^2 E_{n_i} \right) \), where \(|c_i|^2 \) is the probability of each state and \(E_{n_i} \) is the energy of each state.

Explanation:

• Given mixed state:

  • ​ \(\psi = \frac{1}{3} \psi_{1,0,0} + \frac{2}{3} \psi_{2,1,0} + \frac{2}{3} \psi_{3,2,-2} \)

• The probability of each state is given by the square of the coefficient:

  • For \(\psi_{1,0,0} \): \( |c_1|^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \)

  • For \(\psi_{2,1,0}\): \(|c_2|^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}\)

  • For \( \psi_{3,2,-2}\): \(|c_3|^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \)

• Calculate the energy of each state:

  • For ( n = 1 ): \(E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \)

  • For ( n = 2 ): \(E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \)

  • For ( n = 3 ): \(E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV} \)

• Calculate the expectation value of energy:

  • ​\( 〈 E 〉 = \frac{1}{9} \times (-13.6) + \frac{4}{9} \times (-3.4) + \frac{4}{9} \times (-1.51) \)

• Perform the calculations:

  • \( \frac{1}{9} \times (-13.6) = -1.51 \, \text{eV} \)

  • \(\frac{4}{9} \times (-3.4) = -1.51\)

  • \(\frac{4}{9} \times (-1.51) = -0.67 \, \text{eV}\)

• ​Expectation Energy 〈E〉 ≈ -3.7 eV

Conclusion:

 The expectation value of energy of this electron in eV will approximately is -3.7

Concept:

The normalization condition requires that the total probability is equal to 1, so the wavefunction must satisfy the following condition:

Explanation:

• Let the wavefunction be written as:

  • ​\( \psi_- = c(1S_A - 1S_B) \)

• Now, applying the normalization condition:

  • ​\( \int |\psi_-|^2 d\tau = 1 \)

• This gives:

  • ​\( |c|^2 \int (1S_A - 1S_B)^2 d\tau = 1 \)

• Expanding the square:

  • ​\( |c|^2 \left( \int (1S_A^2) d\tau + \int (1S_B^2) d\tau - 2 \int (1S_A 1S_B) d\tau \right) = 1 \)

• Assuming that the wavefunctions are orthonormal:

  • ​\( \int 1S_A^2 d\tau = \int 1S_B^2 d\tau = 1 \text{ and } \int 1S_A 1S_B d\tau = S \)

• Thus, the normalization condition becomes:

  • ​\( |c|^2 (1 + 1 - 2S) = 1 \)

• Solving for c" id="MathJax-Element-54-Frame" role="presentation" style="position: relative;" tabindex="0">cc $c$ :

  • ​\( |c|^2 (2 - 2S) = 1 \)

  • ​\( |c|^2 = \frac{1}{2(1 - S)} \)

• Therefore, the normalized wavefunction is:

  • ​\( \psi_- = \frac{1}{\sqrt{2(1 - S)}}(1S_A - 1S_B) \)

Conclusion:

The normalized wavefunction is \( \frac{1}{\sqrt{2(1-S)}}(1S_A - 1S_B)\) , which corresponds to option 4.

Concept:

In quantum mechanics, the total angular momentum (L) of a particle in a central potential is an important observable. The eigenvalue of the (L²) operator gives the total angular momentum of the system in terms of the quantum number l.

For a wavefunction (\( \psi(r) \)), where the radial part of the wavefunction depends only on r, the angular part corresponds to spherical harmonics (\( Y_{l,m}(\theta, \phi) \)). The eigenvalue of the (L²) operator is:

\(L^2 \psi = l(l+1)\hbar^2 \psi\)

• Wavefunction in Spherical Coordinates: From the handwritten notes, the wavefunction is written as:

  • \( \psi(r) = zf(r), where \ z = r\cos(\theta) \) in spherical coordinates.

    • This implies that l = 1 because the wavefunction has a dependence on \( \cos(\theta) \).

• Total Angular Momentum: Since l = 1, the total angular momentum squared is given by:

  • ​\( L = \sqrt{l(l+1)}\hbar = \sqrt2\hbar\)

Explanation: 

• According to the given wavefunction (\( \psi(r) = zf(r)\) ), the wavefunction depends on (\( \cos(\theta) \)), which corresponds to ( l = 1 ). For this value of ( l ), the total angular momentum squared is:

  • ​\( L^2 = 1(1+1)\hbar^2 = 2\hbar^2\)

Conclusion:

The correct answer is Option 1: (\( 2\hbar^2\) ), based on the value of ( l = 1 ) for the given wavefunction.

Concept:

The virial theorem relates the average kinetic energy and potential energy in a stable system bound by inverse-square forces, such as the hydrogen atom. The main points of the virial theorem are:​

• Application to Hydrogen Atom: In a hydrogen atom, the electron is bound to the nucleus by an inverse-square Coulomb potential, and the virial theorem applies.

• Kinetic Energy and Potential Energy Relation: According to the virial theorem, the mean kinetic energy of the electron 〈T〉 is half of the magnitude of the potential energy 〈V〉, with a negative sign.

• Implication for Total Energy: The total energy 〈E〉 of the system is related to the potential energy as:

  \(⟨E⟩ = ⟨T⟩ + ⟨V⟩ = -\frac{1}{2} ⟨V⟩\)

Explanation: 

• According to the virial theorem, the mean kinetic energy 〈T〉 of an electron in a hydrogen atom is given by

  • ​\(⟨T⟩ = -\frac{1}{2} ⟨V⟩\)

• This relation shows that the kinetic energy is half of the magnitude of the potential energy, but with a negative sign, consistent with the virial theorem for systems bound by inverse-square forces.

Conclusion:

The correct answer is Option 1, which follows from the virial theorem.

The correct answer is \(\frac{19 h^2}{72 m L^2}, g=2 \)

Explanation:-

Unequal sides
first Excited State
The first excited state will be the next combination, and it must be a combination of quantum numbers that is higher than the first excited state. The possible combinations that we need to check are:
(2,1,1)
(1,2,1)
(1,1,2)

So degeneracy g = 3

∈n.x ny n2 = \(\rm \frac{h^2}{8m}\left[\frac{nx^2}{L^2}+\frac{ny^2}{(2L^2)}+\frac{n_x^2}{(3L)^2}\right]\)

\(\rm =\frac{h^2}{8mL^2}\left[n_x^2+\frac{n_y^2}{4}+\frac{n_z^2}{9}\right]\)

\(\rm =\frac{h^2}{8mL^2}\left[\frac{36n_x^2+9n_y^2+4n_z^2}{36}\right]\)

\(\rm =\frac{h^2}{288mL^2}\left[{36n_x^2+9n_y^2+4n_z^2}\right]\)

\(\rm \in_{121} =\frac{h^2}{288mL^2}\left[{36+9(4)+4}\right]\)

\(\rm =\frac{76h^2}{288 mL^2}=\frac{19 h^2}{72mL^2}\)

Conclusion:-

So, The energy and degeneracy g, of the second excited state of a particle of mass m moving freely inside a rectangular parallelopiped of sides L, 2L and 3L, are respectively \(\frac{19 h^2}{72 m L^2}, g=3 \)

The correct answer is \(\frac{5}{2} \hbar \omega\)

Explanation:-

ψ = ψ0(x) + ψ1(x)

ψ0(x) → Ground state

ψ1(x) → Exited state

\(\rm \left(2m+\frac{3}{2}\right)ℏ ω\)

ε0 = 3/2 ℏω , m = 0, 1, 2, 3

ε1 = (2 × 1 + 3/2)ℏω

= 7/2 ℏω

ψ normelized = \(\rm \frac{1}{\sqrt2}\psi_0+\frac{1}{2}\psi _1\)

Σ|Ci|2 = 1

<ε> = ε |Ci|2 ∈1

= |Ci|2 ∈1 + |C2|2 ∈2

= \(\rm \left(\frac{1}{2}\right)^2\frac{3}{2}h\omega+\left(\frac{1}{\sqrt 2}\right)^2\frac{7}{2}h \omega\)

\(\rm =\frac{1}{2}\left(\frac{10}{2}\right)h \omega\)

\(=\frac{5}{2}h\omega\)

Conclusion:-

So, The energy of the particle is \(\frac{5}{2} \hbar \omega\)

The correct answer is 3/2 a0

Concept:-

• Bohr Model & Bohr Radius: While the Bohr model is a simplified view of the atom introduced prior to quantum mechanics, the Bohr radius remains a fundamental constant in describing atomic scales.
• Wavefunction ((ψ)): A mathematical function that describes the quantum state of a particle, including its position in space.
• Probability Density: The square of the wavefunction's magnitude ((|ψ|²)) gives the probability density, which describes how likely it is to find an electron at any given location around the nucleus.

Explanation:- 

The general formula is \(_{n,l} = \frac{a_0}{2}(3n^2-l(l+1))\)

Putting value of n,l for 1s in above equation

\(_{1,0} = \frac{a_0}{2}(3(1)^2-0(0+1))\)

\(_{1,0} = \frac{3a_0}{2}\)

Conclusion:-

So, The average radius for 1s orbital is  3/2 a0

The correct answer is 1.

Concept:- 

• For a particle in a 3D box, the wave function and energy can be represented as below:

\(\Psi_{3D(n_{x},n_{y},n_z)} = \sqrt{\frac{2}{l_{x}}} sin\frac{n_{x}\pi_{x}}{l_{x}}\sqrt{\frac{2}{l_{y}}}sin\frac{n_{y}\pi_{y}}{l_{y}}\sqrt{\frac{2}{l_{z}}}sin\frac{n_{z}\pi_{z}}{l_{z}} \)

and the corresponding energy is:

\(E_{2D}= \left [ \frac{n_{x}^{{2}}}{l_{x}^{{2}}} + \frac{n_{y}^{{2}}}{l_{y}^{{2}}}+ \frac{n_{z}^{{2}}}{l_{z}^{{2}}} \right ] \frac{h^{2}}{8m} \)

• The condition for a 3D cubic box is \(l_x=l_y=l _z=L\)
• Energy for a 3D cubic box will be:

\(E_{2D}= \frac{(n_{x}^{{2}}+n_{y}^{{2}}+n_{z}^{{2}})h^{2}}{8mL^{2}}\)

where: E is the energy, h is Planck's constant, nx, ny, and nz are the quantum numbers associated with the particle (they can be any positive integer), m is the mass of the particle, and L is the length of the box.

CALCULATION:

• Given that the energy is four times the lowest energy level:
  • Lowest energy, \( E_{1,1,1} = \frac{3h^2}{8mL^2} \).
  • Thus, the target energy is \( 4 \times \frac{3h^2}{8mL^2} = \frac{3h^2}{2mL^2} \).
• To achieve this energy level, the quantum numbers must satisfy:
  • \(n_x^2 + n_y^2 + n_z^2 = 6 \).
  • Possible combinations are:
    • (2, 1, 1)
    • (1, 2, 1)
    • (1, 1, 2)

The correct answer is: Option 4 - Degeneracy is 3