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

CONCEPT:

de Broglie Wavelength

• The de Broglie wavelength of a particle is given by the equation:

  λ = h / p

  where h is Planck's constant and p is the momentum of the particle.
• For a particle with kinetic energy E, the momentum p is related to the kinetic energy by:

  p = √(2mE)

  where m is the mass of the particle.
• Thus, the de Broglie wavelength becomes:

  λ = h / √(2mE)

EXPLANATION:

• We are given:
  • Kinetic energy of the electron is E and kinetic energy of the proton is 3E.
  • Mass of the proton is 1836 times that of the electron, i.e., m_p = 1836 m_e.
• For the electron, the de Broglie wavelength is:

  λ_e = h / √(2m_eE)

• λ_p = h / √(2m_p3E) = h / √(2 * 1836 m_e * 3E)

• Taking the ratio of the wavelengths:

  λ_e / λ_p = (h / √(2m_eE)) / (h / √(2 * 1836 m_e * 3E))

  λ_e / λ_p = √(1836 * 3)

  λ_e / λ_p = √(5508) ≈ 74.21

Therefore, the ratio of the de Broglie wavelengths (λ_e / λ_p) is approximately 74.21.

CONCEPT:

Particle in a 3D Cubic Box

• The energy eigenfunctions for a particle in a 3D box with infinite potential walls are given as:

  ψ_nx,ny,nz(x, y, z) = A sin(n_xπx/L) sin(n_yπy/L) sin(n_zπz/L)

• The corresponding energy eigenvalues are:

  E_nx,ny,nz = (h² / 8mL²) × (n_x² + n_y² + n_z²)

• Only those linear combinations of wavefunctions that have the same total quantum number sum (i.e., same energy) can be eigenfunctions of the Hamiltonian.

EXPLANATION:

• ϕ₁ = (1/√2)(ψ_1,4,1 − ψ_2,2,3)
  • Energy of ψ_1,4,1 = 1² + 4² + 1² = 18
  • Energy of ψ_2,2,3 = 2² + 2² + 3² = 17
  • Different energies → Not an eigenfunction
• ϕ₂ = (1/√2)(ψ_1,5,1 + ψ_3,3,3)
  • Energy of ψ_1,5,1 = 1² + 5² + 1² = 27
  • Energy of ψ_3,3,3 = 3² + 3² + 3² = 27
  • Same energy → Valid eigenfunction
• ϕ₃ = (1/√2)(ψ_1,3,8 + ψ_3,8,1)
  • Energy of both = 1² + 3² + 8² = 27
  • Same energy → Valid eigenfunction
• ϕ₄ = (1/2)ψ_3,3,1 + (√3/2)ψ_2,4,1
  • Energy of ψ_3,3,1 = 19, ψ_2,4,1 = 21
  • Different energies → Not an eigenfunction

Therefore, the eigenfunctions of the Hamiltonian operator are: ϕ₂ and ϕ₃ (Options 1 and 3).

CONCEPT:

• (a) Time-independent Schrödinger equation: Can be solved exactly only for hydrogenic atoms (single-electron systems). Be²⁺ is a two-electron system and hence requires approximation methods like Hartree-Fock.
• (b) Particle in a 1D box: The trial wavefunction must satisfy two conditions:
  • Be continuous and single-valued.
  • Satisfy boundary conditions: Ψ(0) = Ψ(l) = 0.
• (c) Fermions: Must obey the Pauli Exclusion Principle and have wavefunctions that are anti-symmetric under exchange of any two fermions.
• (d) Born-Oppenheimer Approximation: Separates electronic motion from nuclear motion, **but not** rotational and vibrational motions directly. It’s used to separate electronic and nuclear degrees of freedom.

EXPLANATION:

• Statement (a): False
  • Be²⁺ is not a one-electron system; it has two electrons. Exact solution is not possible.
• Statement (b): True
  • Given trial function: ϕ(x) = [(3l³)^1/2 × x] does not satisfy Ψ(0) = 0 since at x = 0, the function is zero, but its normalization is problematic. The form also doesn’t vanish at x = l. Thus, it's unacceptable.
• Statement (c): True
  • Fermionic wavefunctions are anti-symmetric under exchange of any two fermions. That’s a fundamental postulate of quantum mechanics.
• Statement (d): False
  • Born-Oppenheimer approximation separates electronic and nuclear motion, not directly vibrational and rotational motions of nuclei.

Therefore, the correct option is Option 3: (a) False, (b) True, (c) True, (d) False.

CONCEPT:

Particle in a 1D Box (Infinite Potential Well)

• In quantum mechanics, a particle in a one-dimensional infinite potential well has discrete energy levels and corresponding wavefunctions.
• For two non-interacting particles:
  • The total energy is the sum of individual energies: E_total = E₁ + E₂
  • The total wavefunction is the product of individual wavefunctions (not the sum): Ψ_total = ψ₁ × ψ₂
• The principle of superposition applies only to linear combinations of eigenfunctions for a **single** particle system, not multiple independent particles.

EXPLANATION:

• Statement 1: The total energy is E₁ + E₂ – Correct
  • This is true for non-interacting systems: total energy is additive.
• Statement 2: The total wavefunction is ψ₁ + ψ₂ – Incorrect
  • For two distinct, non-interacting particles, the correct form is a product: ψ₁ψ₂.
• Statement 3: The total energy is E₁E₂ – Incorrect
  • This is physically meaningless in this context. Energies add, they do not multiply.
• Statement 4: The total wavefunction is ψ₁ψ₂ – Correct
  • This is the standard treatment for two non-interacting particles in quantum mechanics.

Therefore, the incorrect statements are Option 2 and Option 3.

CONCEPT:

Perturbation Theory – Ground State Energy Corrections

• The energy of a perturbed system is given by a power series expansion:

  E = E⁰ + E¹ + E² + ...

• E⁰: zeroth-order (unperturbed) energy
• E¹: first-order correction
• E²: second-order correction

Important Properties (for Ground State):

• E² is always ≤ 0 (second-order correction is usually negative)
• Total corrected energy (E⁰ + E¹) is always greater than or equal to the exact ground state energy:

  E⁰ + E¹ ≥ E₀

1. E⁰ + E¹ + E² > E₀ 
   Not always true, since E² is negative and may reduce the total below E₀.


2. E⁰ + E¹ > 0 
   Irrelevant — total energy being greater than 0 is not a guaranteed property.


3. E⁰ + E¹ ≥ E₀ 
    This is always true — based on the variational principle and behavior of perturbation expansion.


4. E² > 0 
   Incorrect — E² is always negative or zero in the ground state.

The correct answer is \(\rm E_0^0+E_0^1\ge0\)

Explanation:-

Angular Momentum Operators in Quantum Mechanics is a mathematical representation of the angular momentum linked with the physical state of the system. For example, L_x, L_y, L _z.

Calculation:- 

For, Angular Momentum Operators ⇒ [L_x, L_y] = i h L _z ,  [L_y, L _z] = i h L_x ,  [L _z, L_x] = i h L_y.

For Spin Angular Momentum Operators ⇒ [S_x, S_Y] =  i h  S_z,  [S_y, S_z] =  i h  S_x, [S_z, S_x] =  i h  S_y.

 \(⟨α|\hat{S}_x \hat{S}_y - \hat{S}_y \hat{S}_x| α ⟩ = ih^2a\) 

⇒ ˂ a | S_x S_y - S_y S_x | a˃ 

⇒ ˂ a | [ S_x ,S_y ] | a ˃ 

⇒ ˂ a | i h  S_z | a ˃

⇒  i h ˂ a |S_z| a ˃ 

⇒ i h x 1/2 h ˂a | a˃ 

⇒ 1 / 2 i h² = 0.5 i h².

Concept:

• In one dimension, wave functions are often denoted by the symbol ψ(x,t). 
• They are functions of the coordinate x and the time t. But ψ(x,t) is not a real, but a complex function, the Schrodinger equation does not have real, but complex solutions.
• The wave function of a particle, at a particular time, contains all the information that anybody at that time can have about the particle. 
• But the wave function itself has no physical interpretation. It is not measurable. However, the square of the absolute value of the wave function has a physical interpretation. 
• In one dimension, we interpret |ψ(x,t)|2 as a probability density, a probability per unit length of finding the particle at a time t at position x.
• The probability of finding the particle at time t in an interval ∆x about the position x is proportional to |ψ(x,t)|2∆x.
• This interpretation is possible because the square of the magnitude of a complex number is real. 

Explanation:

• For the probability interpretation to make sense, the wave function must satisfy certain conditions. We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one. This leads to the requirements listed below.
• The wave function must be single valued and continuous. The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.
• We must be able to normalize the wave function.  We must be able to choose an arbitrary multiplicative constant in such a way, so that if we sum up all possible values ∑|ψ(xi,t)|2∆xi  we must obtain 1.
• An acceptable wave function may be odd as well as even.

So, the correct options are single valued and continuous.

Concept:

• For a particle in a 2D box, the wavefunction is given by

\(\Psi {n_x},{n_y} = N\sin \left( {{{{n_x}\pi x} \over L}} \right)\sin \left( {{{{n_y}\pi y} \over L}} \right)\), where N is the normalization constant.

• The energy of a particle in a 2D box of side length l is given by,

\(E = {{({n_x}^2 + {n_y}^2){h^2}} \over {8m{l^2}}}\), where n_x and n_y is the quantum number.

Explanation:

• The three non-interacting electrons can be arranged in a 2D square box as follows:

(1, 2) (2, 1) 1 - \(\rm E_1=1\times5\frac{h^2}{8ml^2}\)

(1, 1) \(\rm E_2=2\times\frac{2h^2}{8ml^2}\)

\(\rm E_T=\frac{9h^2}{8ml^2}=E_1+E_2\)

• In the ground state (g.s), two electrons can be placed in a singly degenerated energy state (1,1). The energy is given by,

\({E_1} = 2 \times {{(1 + 1){h^2}} \over {8m{l^2}}}\)

\( = {{4{h^2}} \over {8m{l^2}}}\)

• ​In the first exited state, one electron can be placed in a doubly degenerated energy level (2,1) and (1,2). The energy is given by,

​\({E_2} = {{({2^2} + 1){h^2}} \over {8m{l^2}}}\)

\( = {{5{h^2}} \over {8m{l^2}}}\)

The energy of the system is thus,

\( = {{4{h^2}} \over {8m{l^2}}} + {{5{h^2}} \over {8m{l^2}}}\)

\( = {{9{h^2}} \over {8m{l^2}}}\)

Conclusion:

Hence, the ground state energy in units of \(\left( {\frac{h^2}{{8mL^2}}}\right)\) is 9

Explanation:​

de Broglie wavelength of electrons:

•  Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well,i.e. matter has dual nature.
• The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength (λ) of electrons can be calculated from Plancks constant h divided by the momentum of the particle
• So, according to de Broglie, every object has a dual nature- a particle and a wave nature whose wavelength is given by
  \(\lambda = \frac{h}{{mv}}\)
  or \(\lambda = {h\over p};p=mv\)
  where m is the mass of the particle,v is the velocity of the particles and h is the Planck's constant.​
• The wave nature is predominant in particles of small mass like electrons and negligible in bodies of large masses.
• Waves are dispersed over a wide region and not confined.

Calculation:

Given:

• The speed of Argon atom =  250 m s−1 
• h = 6.626 × 10−34 kg m2 s −1
• Mass of Argon = 40 amu = 40 g/mol
• ​The wavelength of argon =? in pm.

NOTE: We have to find the wavelength in picometers, so we have to convert everything into the same units before calculation. So, g is converted to kg.

• Hence, the wavelength =
  \(\lambda = \rm \frac{{6.626{\rm{ }} × \;10^{ - 34} \ kg m^2 s ^{−1}}}{{250m/s × 40 × 10^ {- 3} \ kgmol^{-1}}}\)

Now, for converting moles to numbers, we have to divide the atomic mass of Argon by Avogadro's number, so the equation becomes:

 \(\rm \lambda = \frac{{6.626{\rm{ }} × \;10^{ - 34} \ kg m^2 \ s ^{−1}6.022 × 10^{23}mol^{-1}}}{{250m/s × 40 × 10^ {- 3} \ kgmol^{-1}}}\)

The value of wavelength now comes as: 00399 × 10^-8 m = 39.9 × 10¹² m = 39.9 pm.

Hence, the wavelength of the argon atom is 39.9 pm.

Concept:

• Average Energy/ Expectation value of energy of an electron is the sum of expectation value of kinetics and potential energy and it is written as

\(=+ \)

\(where,\\= Total\;energy\)

\(= Avg.\;kinetic \;energy\)

\(= Avg.\;potential\;energy\)

• Also, expectation value of kinetic energy of electron is related to mass of electron (m_e) and the velocity (v) as:

\(=\frac{1}{2}m_e\)

Explanation:

According to the question, the potential energy of the particle/electron inside the well is 0. So, it can be concluded the given energy value is the kinetic energy of the electron

\(=\frac{1}{2}m_e+0\)

\(=\frac{1}{2}m_e\)

The above relation can be rearranged to get the expectation value of square of velocity as follows:

\(=\frac{2}{m_e}\;\)       ......relation(1)

given,

\(=13.6eV=13.6\times1.6\times10^{-19}J \)

\(m_e=9.1\times10^{-31}Kg \)

Equating the values in relation(1), will give:

\(= 2\times\frac{13.6\times1.6\times10^{-19}J}{9.1\times10^{-31}Kg}\;\)

\(=4.7\times10^{12}m^2s^{-2} \)

Conclusion:

Hence, the expectation value of square of velocity of electron confined in the infinite potential well is \(4.7\times10^{12}m^2s^{-2} \)

Concept:

For linear and position momentum:

[x̂, p̂_x] = iℏ   ....(1)

[p̂_x, x̂] = -iℏ   ...(2)

[x̂ⁿ, p̂_x] = nx^n-1 [x, p̂_x]   ....(3)

[p̂_xⁿ, x̂] = np_x^n-1[p̂_x, x̂] .....(4)

Explanation:

Given → [x, p_x²]

→ Now using equation (4) and (1),

[x, p_x²] = 2 \(p_x^{2-1}\) [x, p_x]

= 2 p_x iℏ

Conclusion:-

∴ option '2' is correct i.e. [x, px2] = 2iℏp_x

Concept:-

A 1-dimensional (1D) box, also known as a particle in a one-dimensional box or a particle in a box, is a simple quantum mechanical model that describes the behavior of a particle confined to move within a one-dimensional region, typically between two walls or barriers. This model is often used in introductory quantum mechanics courses to illustrate the quantization of energy levels and wavefunctions in a confined system.

Given:

\({h^2 \over 8m} = k\)

The average bond length of C-C = 1.5 Å

Explanation:-

Bond length for the whole conjugated system having 10 bonds are

= 1.5 x10 

=15 Å

We know that, 

\(∆E=hν={(∆n)^2}{h^2 \over 8ma^2}\)

\(∆E={(∆n)^2}{k \over 15^2}\)

In the given conjugated system 10π electrons are arranged in the following manner -

The 10 π electrons make the ground state and the first transition state occurs at minimum transition level at n= 6.

\(∆E={(6^2 -5^2)}{k \over 225}\)

\(∆E={11k \over 225}\)

• Now, the frequency required to cause a transition from the ground state of the system to the first excited state is,

\(∆E=hν={11k \over 225}\)

\(ν={11k \over 225 h}\)

Conclusion:-

Therefore, the frequency of radiation required to cause a transition from the ground state to the first excited state of the system is \(\frac{11\,k}{225\,h}\).

Concept:

In a three-dimensional cubic box, the energy of a particle is given by the equation:

\(E = h²(n₁² + n₂² + n₃²) / 8mL²\)

where: E is the energy, h is Planck's constant, n₁, n₂, and n₃ 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.

Explanation:

You're given that the energy E is 27h²/ 8mL² . Setting this equal to the energy equation, we get:

\(27h² / 8mL² = h²(n₁² + n₂² + n₃²) / 8mL²\)

Solving for n₁² + n₂² + n₃², we find that it equals 27. This means that the sum of the squares of the three quantum numbers equals 27.

The possible sets of quantum numbers (n₁, n₂, n₃) that satisfy this equation are (3, 3, 3), (1, 1, 5), (1, 5, 1), and (5, 1, 1). Each set of quantum numbers corresponds to a different state of the particle, so there are four states that have this energy.

Conclusion:-

Therefore, the degeneracy of the state with energy \(27h² / 8mL²\) is 4

Concept:

• The average value in Quantum Mechanics is the average result obtained when a large number of measurements are made on identical systems or systems in the same state.
• For an observable quantity A, the average value or expectation value is given by,

\(\left\langle {\rm{A}} \right\rangle {\rm{ = }}\int {{{\rm{\Psi }}^{\rm{*}}}{\rm{A\Psi d\tau }}} \)

Explanation:

• The normalized ground state wave function for a one-electron system like an H atom (He⁺) is:

\({\Psi _{1s}} = N{e^{ - {r \over {Z{a_ \circ }}}}}\)  where, N is the normalization constant and Z is the atomic number of the nuclei.

• The average value or expectation value of r is given by,

\(\left\langle r \right\rangle = \int {{\Psi ^ * }_nr{\Psi _n}} d\tau \)

• Now, for an electron in 1s orbital of He+, the average value of r, (r) is given by,

\(\left\langle r \right\rangle = \int {{\Psi ^ * }_{1s}r{\Psi _{1s}}} d\tau \), where \(d\tau = {r^2}dr\sin \theta d\theta d\phi \)

\( = N\int_0^ \propto {r3{e^{ - {{2r} \over {Z{a_ \circ }}}}}} dr\int_0^\pi {sin\theta d\theta } \int_0^{2\pi } {d\phi } \)

\( = {{3{a_ \circ }} \over {2Z}}\)

• Now, for the He+ ion the atomic number

(Z) = 2.

• Thus, the average value of r, \(\left\langle r \right\rangle \)  for He+ ion is

​\( = {3 \over 4}{a_ \circ }\).

Conclusion:

Hence, for an electron in 1s orbital of He+, the average value of r, (r) is \({3 \over 4}{a_ \circ }\)

CONCEPT:

Commutator Relations Between Total Angular Momentum and Raising/Lowering Operators

• In quantum mechanics, the raising (\(L_+\)) and lowering (\(L_-\)) operators change the magnetic quantum number m, but they do not affect the total angular momentum \(L^2\).
• The raising operator \(L_+\) increases the magnetic quantum number m by 1, and the lowering operator \(L_-\) decreases it by 1.
• The commutator between the total angular momentum \(L^2\) and these raising and lowering operators is zero because applying these operators does not change the total angular momentum magnitude \(L^2\).

EXPLANATION:

• The total angular momentum operator \(L^2\) represents the magnitude of the angular momentum vector, while the raising and lowering operators only affect the projection along the z-axis (i.e.,\( L_z\)).
• The commutator of \(L^2\) with the raising and lowering operators is zero because they do not affect the magnitude of the total angular momentum. Instead, they modify only the \(L_z\) component:
  • \([L^2, L_+] = 0\) \([L^2, L_-] = 0\)

CALCULATION:

• The total angular momentum operator \(L^2\) is defined as:
  • \(L^2 = L_x^2 + L_y^2 + L_z^2\)
• The raising and lowering operators are defined in terms of the angular momentum components \(L_x\) and \(L_y\):
  • \(L_+ = L_x + iL_y\) \(L_- = L_x - iL_y\)
• Now, we calculate the commutator \([L^2, L_+]\) by expanding it:
  • \([L^2, L_+] = [L_x^2 + L_y^2 + L_z^2, L_x + iL_y]\)
• This can be broken into three parts:
  • \([L_x^2, L_+] = 0\) \([L_y^2, L_+] = 0\) \([L_z^2, L_+] = 0\)
• Therefore, the total commutator \([L^2, L_+]\) results in:
  • \([L^2, L_+] = 0\)
• Similarly, for the lowering operator \(L_-\), we follow the same steps:
  • Breaking this into parts also gives zero for each commutator:\([L^2, L_-] = [L_x^2 + L_y^2 + L_z^2, L_x - iL_y]\)
    • \([L_x^2, L_-] = 0\) \([L_y^2, L_-] = 0\) \([L_z^2, L_-] = 0\)
• Thus, the final commutator is:
  • \([L^2, L_-] = 0\)

CONCLUSION:

The correct commutator relation is: \( [L^2, L_+] = [L^2, L_-] = 0\).