Application Of Schrodinger Wave Equation – MSQ - Physics MCQ

Case 1: Infinite Potential Well

1. In an infinite potential well, the wavefunction inside the well is described by:

   

   This is due to the boundary conditions that the wavefunction must vanish at the walls (V → ∞).

2. Outside the infinite potential well, the wavefunction ψoutside\psi_{\text{outside}}ψoutside​ is zero because the particle cannot exist outside the well (V = ∞).

Case 2: Finite Potential Well

1. For a finite potential well, the wavefunction inside the well is described by a combination of sinusoidal terms, such as:

   ψ_inside = Bcos(kx)

   where ​​, and E < V₀​.

2. Outside the well, the potential V(x) = V₀​ leads to an exponential decay of the wavefunction since E < V₀​:

   ψ_outside = Be^±kx

   where

Correct Options:

• Option B: Correctly describes the wavefunction inside the infinite well and inside the finite well (Bcos(kx)).
• Option C: Correctly describes the wavefunction outside the infinite well (ψ_outside ​= 0) and outside the finite well (Be^±kx).

Incorrect Options:

• Option A: Incorrect as it implies a form of the wavefunction for the finite well that doesn't match the expected solutions.
• Option D: Incorrect as it assumes ψ_outside = Bcos(kx) for the finite well, which is not valid for decaying solutions.

Thus, the answer is B, C.

The steady-state form of the Schrödinger wave equation refers to the time-independent Schrödinger equation, which describes the wavefunction for a particle in a potential field.

1. The equation is linear because the principle of superposition applies. This means that if two solutions exist, their linear combination is also a solution.
2. It is a second-order differential equation but is not quadratic or derivable in the context of mathematical classification.
3. Therefore, the equation is best described as linear in its form.

The wavefunction ψ(x, t) is given as:

Here, ψ(x) is the spatial part of the wavefunction, and e^−iEt​/h represents the time-dependent phase factor.

Analysis of the Options:

• Option A: ∣ψ(x)∣² is the probability density in space and is independent of time since . Similarly, , which also has no time dependence. Thus, this option is correct.

• Option B: If ψ(x) is normalized then the time-dependent wavefunction ψ(x, t) remains normalized because the time-dependent factor e^−iEt​/h does not affect the normalization . Thus, this option is correct.

• Option C: ∣ψ(x)∣² is the spatial probability density, which is independent of time. ∣ψ(x,t)∣²=∣ψ(x)∣², so it is also independent of time. Thus, this option is incorrect.

• Option D: If ψ(x) is normalized, ψ(x, t) will always be normalized due to the time factor not affecting the probability density. Thus, this option is incorrect.

The solution of a symmetric infinite potential well consists of 

Now, any function can be written as a linear combination of these functions (sin and cos)

∴ They form a complete let

The correct answers are: They are alternately even and odd with respect to the centre of the well, As we go up in energy, each successive state has one more node. i.e.  ψ₁(x) has none, ψ₂(x) has one, ψ₃(x) has 2 and so on, They are mutually orthogonal, They form a complete set

The correct answers are: 

1. Energy Levels for an Infinite Potential Well: In a one-dimensional infinite potential well, the energy levels are quantized and given by:

   

   where:

   • nnn is the quantum number (n = 1, 2, 3,…),
   • h is Planck's constant,
   • m is the particle's mass,
   • L is the width of the well.

2. Energy Levels for a Finite Potential Well: In a finite potential well, the energy levels are also quantized but slightly lower than those in the infinite potential well due to the finite height of the potential, which allows tunneling effects. However, the functional form remains approximately the same for lower energy levels:

   

   for sufficiently deep wells.

3. Harmonic Oscillator Energy Levels: The energy levels given in Option A:

   

   represent the quantized energy levels for a quantum harmonic oscillator, not a potential well. Hence, Option A is incorrect.

Analysis of Options:

• Option A: Incorrect, as this describes the energy levels of a quantum harmonic oscillator, not a one-dimensional potential well.
• Option B: Correct, because the energy levels for both the infinite and finite potential wells are approximately given by ​, especially for deep finite wells.
• Option C: Incorrect, as the energy levels are approximately the same in form for both cases, differing only due to tunneling effects in the finite case.
• Option D: Incorrect, as Option B is correct.

Probability is 0 at x = 0 & a
For 

Probability density is zero at x = 0, a/2 , a

Probability density zero at x = 0, 

The correct answers are: 

The Schrodinger equation solution are 

∴  momentum eigenfunction.

The correct answers are: The Schrodinger equation yields the solution to be  (linear combination of the two), The solutions of the Schrodinger equation are both energy and momentum eigenfunctions

 for a harmonic oscillator

 

∴   evenly spaced.

The correct answers are: A spectrum of evenly spaced energy levels, A non-zero probability of finding the oscillator outside the classical turning points.

Concept:

• Schrodinger wave equation is a mathematical expression that describes the energy and position of the electron in space and time, taking into account the matter wave nature of the electron inside an atom.

• It is based on three considerations. They are:

  1. Classical plane wave equation, (wave which satisfies 
  2. Broglie’s Hypothesis of matter-wave, (the wavelength of matter-wave is inversely proportional to the linear momentum.)
  3. Conservation of Energy. (total energy = kinetic + potential)

Schrodinger equation gives us a detailed account of the form of the wave functions or probability waves that control the motion of some smaller particles.

The equation also describes how these waves are influenced by external factors.

Moreover, the equation makes use of the energy conservation concept that offers details about the behavior of an electron that is attached to the nucleus.

Explanation:

1. For a wave to be a solution of Schrodinger's equation it must satisfy the following conditions:
   1. The wave function must be a single value.
   2. The wave function must be continuous.
   3. The wave function must be finite.
   4. The wave function must be differentiable at every point in space.
      Hence the correct option is 3.