Particle-in-a-box MCQ Quiz in తెలుగు - Objective Question with Answer for Particle-in-a-box - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

CONCEPT:

Energy Levels in a 1D Particle-in-a-Box System and Variational Method

• For a particle confined in a 1D box (potential energy ( V = 0 ) inside the box and ( \(V = \infty \)) outside), the energy levels are quantized, with the ground state energy denoted as ( E₀).
• The expectation value of the Hamiltonian with a trial wave function provides an estimate of the energy level, which is always greater than or equal to the ground state energy ( E₀ ) due to the variational principle.
• Using a linear combination of functions as a trial wave function generally yields a better approximation to the ground state energy, and the resulting energy ( E₂) is expected to be closer to ( E₀) than a single function estimate ( E₁).

CALCULATION:

• Given that the ground state energy of the particle in a 1D box is (\( E_0 = 0.15 \, \frac{\hbar^2}{mL^2} \)).
• The trial wave function ( \(\psi(x) = x(x - L) \)) provides an estimated energy \(\psi(x) = x(x - L) \) that is typically greater than ( E₀) due to the variational principle.
• Using a combination of  x(x - L)  and \( x^2(x - L)^2 \) yields a refined estimate ( E₂), which is closer to ( E₀) but still greater than or equal to it.

CONCLUSION:

The correct relationship between the energies is:

• Option 2: ( \(E_0 < E_2 < E_1\) )

Concept:

In quantum mechanics, an electron in a one-dimensional box (also known as a particle in a box) can only occupy specific energy levels. The energy of an electron in a 1D box of length ( L ) is quantized and given by the formula:

\(E_n = \frac{n^2 h^2}{8mL^2} \)

• Here, ( n ) is the quantum number (n = 1, 2, 3, ...), ( h ) is Planck's constant, ( m ) is the mass of the electron, and ( L ) is the length of the box.
• The wavefunction, ψ_n(x) ), describes the probability amplitude of finding the electron at position ( x ) within the box. It is given by:

  • \(ψ_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right) \)

• The probability density, ( |ψ_n(x)|² ), which determines the likelihood of finding the electron at a particular position ( x ), is obtained by squaring the wavefunction:

  • \(|ψ_n(x)|^2 = \frac{2}{L} \sin^2\left(\frac{n \pi x}{L}\right) \)

Explanation: 

• Given energy = \(\frac{h^2}{2mL^2}\)

• On comparing the energy level with the formula for energy of particle in a box

  • n = 2.

• In the ground state (n = 1), the wavefunction has no internal nodes (points where probability density is zero) within the box; it only has nodes at the boundaries, ( x = 0 ) and ( x = L ). This means the probability of finding the electron is highest at the center of the box (x = L/2) and zero at the edges.

• For the first excited state (n = 2), the wavefunction has one internal node, which occurs at ( x = L/2 ). At this point, the sine function in ψ(x) goes to zero, making the probability density zero as well. This results in a minimum probability of finding the electron at ( x = L/2 ) in the first excited state.

  • ​

• Since the energy level given in the question corresponds to the first excited state (( n = 2 )), we deduce that the probability density has a minimum at ( x = L/2 ). This is the position within the box where the electron is least likely to be found.

Conclusion:

The correct option is: L/2

Concept:

In a 3D cubic box, the energy levels for a particle are quantized and depend on the quantum numbers ( nx ), ( ny ), and ( nz ) in each spatial direction. The energy of a particle in a cubic box of length ( a ) is given by:

\(E = \frac{h^2}{8ma^2} (n_x^2 + n_y^2 + n_z^2) \)

The degeneracy of an energy level is the number of distinct sets of ( (n_x, n_y, n_z) ) that yield the same energy. 

Explanation:

• Given energy:

  • ​ \(E = \frac{7h^2}{4ma^2} \)

• We know that \(E = \frac{h^2}{8ma^2} (n_x^2 + n_y^2 + n_z^2) \), so we equate:

  • \( \frac{h^2}{8ma^2} (n_x^2 + n_y^2 + n_z^2) = \frac{7h^2}{4ma^2} \)

• Simplify by multiplying both sides by \(\frac{8ma^2}{h^2} \):

  • \( n_x^2 + n_y^2 + n_z^2 = 14 \)

• Now, we find the combinations of ( nx ), ( ny ), and ( nz ) that satisfy \(n_x^2 + n_y^2 + n_z^2 = 14\):

  • (nx, ny, nz) = (3, 2, 1)
  • (nx, ny, nz) = (3, 1, 2) 
  • (nx, ny, nz) = (2, 3, 1) 
  • (nx, ny, nz) = (2, 1, 3) 
  • (nx, ny, nz) = (1, 3, 2) 
  • (nx, ny, nz) = (1, 2, 3)

• There are 6 distinct combinations that satisfy this condition, so the degeneracy of the energy level \(E = \frac{7h^2}{4ma^2}\)is 6.

Conclusion:

The degeneracy of the energy level \( E = \frac{7h^2}{4ma^2}\) for a particle in a 3D cubic box is: 6

Correct Option: 4

Concept:

A particle in a 3D box is a quantum mechanical system where a particle is confined to a cubical box with impenetrable walls. The energy levels of the particle are quantized, depending on the quantum numbers along each dimension.

• Energy Levels: The energy for a particle in a 3D box of length a" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">aa $a$ is given by:

  • \(E = \frac{h^2}{8ma^2} (n_x^2 + n_y^2 + n_z^2) \), where nx" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">nxnx $n_x$ , ny" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">nyny $n_y$ , and nz" id="MathJax-Element-6-Frame" role="presentation" style="position: relative;" tabindex="0">nznz $n_z$ are quantum numbers corresponding to each dimension.

• Types of Degeneracy: In a 3D box, degeneracy occurs when different sets of quantum numbers result in the same energy level. This can be classified as:

  • Simple Degeneracy: Caused by symmetry of the system, where multiple quantum states have the same energy.
  • Accidental Degeneracy: Arises when distinct combinations of quantum numbers give the same energy by coincidence, not due to symmetry.

Explanation:

• For a particle in a 3D box, the lowest energy level with accidental degeneracy occurs when multiple quantum number combinations give the same energy.

• At the energy level \(E = \frac{3h^2}{2ma^2} \),

  • the combinations (nx,ny,nz)=(1,1,2)" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">(nx,ny,nz)=(1,1,2)(nx,ny,nz)=(1,1,2) $(n_x, n_y, n_z) = (1,1,2)$ , (1,2,1)" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">(1,2,1)(1,2,1) $(1,2,1)$ , and (2,1,1)" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">(2,1,1)(2,1,1) $(2,1,1)$ produce the same energy, demonstrating accidental degeneracy.

• Other energy levels do not exhibit this form of degeneracy at the lowest levels, confirming that the correct answer is option 3.

Conclusion:

The lowest energy level that is an example of accidental degeneracy for a particle confined in a 3D cubic box of length a" id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">aa $a$ is: 3h22ma2" id="MathJax-Element-11-Frame" role="presentation" style="position: relative;" tabindex="0">3h22ma23h22ma2 $\frac{3h^2}{2ma^2}$

Correct Option: 4

Concept:

Particle in a Box (Quantum Mechanics)

• The wave function of a particle in a box describes the behavior of the particle in a one-dimensional box with infinitely high walls at both ends.
• The quantum number (n) is a positive integer (1, 2, 3, ...) that quantizes the energy levels of the particle in the box.
• The number of nodes (points where the wave function equals zero) in the wave function is directly related to the quantum number.
• The wave function shown corresponds to a specific quantum state, where the number of half-wavelengths in the box is equal to the quantum number n.
• If the wave function shows two nodes (excluding the boundary points), the quantum number n is 3, because the wave function has 3 half-wavelengths between the walls.

Explanation:

• Number of nodes = n - 1
• 2 = n - 1
• n = 3

Therefore, the quantum number for the given wave function is 3 (Option 3).

Solution (3)

In one dimensional box, 

If potential energy v(x) = 0 for L > x > 0

where, L is length.

Potential energy for x ≥ L and x ≤ 0

v(x) = ∞ 

Then, acceptable wave function remains consistent with the boundary condition (A, B, C and D are constant)

= Cx³ (x - L)

Concept:

Particle in a One-Dimensional Box

• The particle in a one-dimensional box is a standard problem in quantum mechanics where a particle is confined to move along a one-dimensional region of space (0 ≤ x ≤ L) with infinite potential walls outside this region (V(x) = ∞ for x < 0 or x > L, and V(x) = 0 inside the box).
• In this scenario, the Schrödinger equation describes the behavior of the particle, and the wavefunction solutions are subject to boundary conditions where the wavefunction must be zero at the walls of the box (x = 0 and x = L).
• The wavefunctions are given by eigenfunctions of the Hamiltonian operator, and the corresponding eigenvalues represent the allowed energy levels of the particle.

Explanation:

For a particle in a one-dimensional box, the eigenfunctions \(( \psi_n(x) )\) are solutions to the time-independent Schrödinger equation with the boundary conditions\( \psi(0) = 0 \ and \ \psi(L) = 0 .\) The general form of the eigenfunctions is given by:

\( \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\)

\(where \ n = 1, 2, 3, \dots\)

The energy levels of the particle are quantized and are given by:

\(E_n = \frac{n^2 h^2}{2mL^2}\)

where n is the quantum number and h is the reduced Planck's constant.

For each eigenfunction, there are specific internal nodes where the wavefunction goes to zero within the box. These nodes correspond to values of x where

\(\sin\left(\frac{n\pi x}{L}\right) = 0\) , i.e., at: \( x = \frac{mL}{n} \)

where m is an integer such that 0 < m < n.

Therefore, the locations of the internal nodes are at \( x = \frac{mL}{n} \)

CONCEPT:

Energy Differences in a Particle Confined in 1-D, 2-D, and 3-D Boxes

• For a particle in a 1-dimensional box of length L , the energy levels are given by:\( E_n^{(1D)} = \frac{n^2 h^2}{8mL^2}\)
  where n is the principal quantum number, h is Planck's constant, and m is the mass of the particle.
• For the 1-D box:
  \( E_{n=2}^{(1D)} = \frac{4h^2}{8mL^2} = \frac{h^2}{2mL^2} \\ E_{n=1}^{(1D)} = \frac{h^2}{8mL^2} \\ \Delta E_1 = E_{2} - E_{1} = \frac{h^2}{2mL^2} - \frac{h^2}{8mL^2} = \frac{3h^2}{8mL^2}\)
• For a particle in a 2-dimensional square box of length L , the energy levels are given by:
  \( E_{n_x, n_y}^{(2D)} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2)\)
  where n_x and n_y are the quantum numbers in the x and y directions, respectively.
• For the 2-D box:
  \( E_{2,1}^{(2D)} = \frac{h^2}{8mL^2}(2^2 + 1^2) = \frac{h^2}{8mL^2}(4 + 1) = \frac{5h^2}{8mL^2} \\ E_{1,1}^{(2D)} = \frac{h^2}{8mL^2}(1^2 + 1^2) = \frac{h^2}{8mL^2}(1 + 1) = \frac{2h^2}{8mL^2} = \frac{h^2}{4mL^2} \\ \Delta E_2 = E_{2,1} - E_{1,1} = \frac{5h^2}{8mL^2} - \frac{2h^2}{8mL^2} = \frac{3h^2}{8mL^2}\)
• For a particle in a 3-dimensional cubic box of length L , the energy levels are given by: 
  \(E_{n_x, n_y, n_z}^{(3D)} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)\)
  where n_x , n_y , and n_z are the quantum numbers in the x, y, and z directions, respectively.
• For the 3-D box:
• \( E_{2,1,1}^{(3D)} = \frac{h^2}{8mL^2}(2^2 + 1^2 + 1^2) = \frac{h^2}{8mL^2}(4 + 1 + 1) = \frac{6h^2}{8mL^2} = \frac{3h^2}{4mL^2} \\ E_{1,1,1}^{(3D)} = \frac{h^2}{8mL^2}(1^2 + 1^2 + 1^2) = \frac{h^2}{8mL^2}(1 + 1 + 1) = \frac{3h^2}{8mL^2} \\ \Delta E_3 = E_{2,1,1} - E_{1,1,1} = \frac{3h^2}{4mL^2} - \frac{3h^2}{8mL^2} = \frac{3h^2}{8mL^2}\)

EXPLANATION:

• \( \Delta E_1 = \Delta E_2 = \Delta E_3 = \frac{3h^2}{8mL^2}\)
• Therefore, the correct relation between the energy differences  \(\Delta E_1 , \Delta E_2 , and \Delta E_3 \) for the three cases is:
  • \(\Delta E_1 = \Delta E_2 = \Delta E_3\)

Therefore, the correct option is Δ1 = Δ2 = Δ3

CONCEPT:

Energy States in a One-Dimensional Box

• In quantum mechanics, a particle in a one-dimensional box (also known as an infinite potential well) has discrete energy levels determined by its quantum number n. This model is often used to illustrate quantum confinement.
• The wave function ψ_n(x) describes the probability amplitude of finding the particle at position x, and n must be a positive integer (n = 1, 2, 3, ...).
• The corresponding energy levels are given by:

  E_n = (ħ²π²n²)/(2mL²), where ħ is the reduced Planck constant, m is the particle's mass, and L is the length of the box.

EXPLANATION:

• The quantum number n = 0 is not allowed because:

  ψ_n(x) = √(2/L) sin(nπx/L)

  • If n = 0, then ψ₀(x) = √(2/L) sin(0) = 0.
  • This implies that the wave function is zero everywhere inside the box.
  • A wave function that is zero everywhere means that the probability of finding the particle within the box is zero, which is not possible for a physically existing particle.

Therefore, the correct answer is the wave function becomes zero everywhere.

The Correct Answer is \(\frac{h^2}{8 m}\left(\frac{2}{a^2}\right), 1 \)

Concept:-

Particle in the Box:

Imagine a particle, often an electron, confined to move along a one-dimensional line within a box. This box has infinite potential energy barriers at its edges, meaning the particle cannot exist outside the box. Within the box, the particle's kinetic energy is the only energy it possesses.

Assumptions:

• The potential energy inside the box is zero.
• The potential energy outside the box is infinite.
• The particle does not experience any forces within the box.

Energy of the Particle in the Box:

In 1D box- \(E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \) where

E_n​  is the energy of the particle in the nth state.
n is a positive integer (1, 2, 3, ...).
ℏ is the reduced Planck constant.
m is the mass of the particle.
L is the length of the box.

In 2D box- \(E_{n_x, n_y} = \frac{h^2}{8m} \left( \frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} \right) \) where

\(E_{n_{x}},_{n_{y}}\) is the energy of the particle in the 
n_x,n_y state.
h is the Planck constant.
m is the mass of the particle.
n_x,n_y are quantum numbers representing the number of nodes in the x and y directions, respectively.
a,b are the lengths of the sides. If, a = b, box is square. If \(a \neq b \)  box is rectangular. 

Explanation:-

\(E_{n_x, n_y} = \frac{h^2}{8m} \left( \frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} \right) \)

For a = 2a and b = a,

\(E_{n_x, n_y} = \frac{h^2}{8m} \left( \frac{n_x^2}{4a^2} + \frac{n_y^2}{a^2} \right) \)

For ground state, nx = 1 and ny = 1

For 1st excited state, nx = 2 and ny = 1

\(E_{2,1} = \frac{h^2}{8m} \left( \frac{2^2}{4a^2} + \frac{1^2}{a^2} \right) \)

\(E_{2,1}=\frac{h^2}{8 m}\left(\frac{2}{a^2}\right)\)

Degeneracy = 1, Since there is no other state with same energy value.

Conclusion:-

the energy and degeneracy of the first excited state, respectively, are \(\frac{h^2}{8 m}\left(\frac{2}{a^2}\right), 1 \)