Application Of Schrodinger Equation MCQ - Physics MCQ

The wave function of a particle in a one-dimensional box of length LLL for the nnn-th state is given by:

For the second excited state (3n = 3), the probability of finding the particle between x = 0 and x = 0.5L is calculated by integrating the square of the wave function over this region:

Substituting ψ_3​(x), we get:

Simplify:

Using the trigonometric identity ​, the integral becomes:

Separate the terms:

Evaluate the first integral:

For the second integral, use the formula for the integral of cos(kx):

Here, a = 6π​/L, so:

At x = 0.5L:

At x = 0:

sin(0) = 0

Thus, the second integral evaluates to 000.

Now, substitute back:

= 1/2

A wave function must satisfy the following key properties:

1. Single-valued: The wave function must have a unique value for every point in space.
2. Continuous and finite: It must not have any discontinuities or infinite values.
3. Normalizable: The integral of the squared wave function over all space must converge to a finite value.

Now, analyzing the options:

• Option A: tanx
  The tangent function has discontinuities at x = π/2, 3π/2,…, making it invalid as a wave function.

• Option B: cotx
  The cotangent function has discontinuities at x = 0, π, 2π,…, so it cannot represent a wave function.

• Option C: secx
  The secant function has discontinuities at x = π/2, 3π/2,…, disqualifying it as a wave function.

• Option D: sinx
  The sine function is continuous, finite, and single-valued across all points. It satisfies the criteria for a valid wave function.

For particle in a box

For ψ₁(x), probability density does not vanish in the middle
for

 

probability density vanishes in the middle

Similarly, it does not vanish for  vanishes for  in the middle of the well.

The correct answer is: states of even n(n = 2, 4...)

The probability distributions for the quantum states of the oscillator without the barrier.

Infinite barrier at the origin means a node at origin, or the wave function goes to zero at x = 0.

By symmetry, the ground state will disappear, as well all the even states.  Odd values remain

The energy of a quantum wave particle is related to its angular frequency (ω\omegaω) by the following formula:

E = ℏω

Where:

• ℏ is the reduced Planck's constant (ℏ = h / 2π).
• ω is the angular frequency of the wave.

This formula represents the quantized energy levels of a particle and is derived from the principles of quantum mechanics.

Momentum Relation:

For reference, the momentum p of the particle is given by:

p = ℏk

Where:

• k is the wave number.

Solving the Schrodinger equation for the region from a to b.

So, in the region from x = a to b, the wave function are in the form of exponential rise/fall.

Given E_n​ = 2 eV = 2 × 1.602×10⁻¹⁹J, we solve for n:

Substitute the values:

Taking the square root:

~ 10⁷

The correct answer is: 6 MeV

Wavefunction for n=3n = 3n=3:
The wavefunction is

:

Expectation value of xxx:
For all stationary states in an infinite potential well, the wavefunction is symmetric or antisymmetric around x = a / 2.

Hence, ⟨x⟩ = a / 2.

Expectation value of x²:

• This integral evaluates to a² / 4 for n = 3.

• Expectation value of p²:
  From the energy eigenvalue relation:
  

Wavefunction for the Ground State:

• The ground state wavefunction for an infinite potential well is:
  
• The corresponding probability density is:
  

This is maximum at x = a/2 because sin² reaches its peak value (1) at the center of the well.

Expectation Value of Position (⟨x⟩):

• The wavefunction is symmetric about x = a/2, so the expectation value of position is:
  

• Hence, it is not zero.

• Energy Eigenvalue of the Ground State:

  The energy eigenvalue for the ground state (n=1) is
  

• The energy is inversely proportional to a², as larger aaa allows the particle to occupy lower-energy states.

• Expectation Value of Kinetic Energy:

  The expectation value of kinetic energy is nonzero because the particle is confined, and its wavefunction has a curvature (derivative). For the ground state: