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Consider a set of wave functions ψi(x). Which of the following condition guarantees that the functions are normalized and mutually orthogonal? (i, j take values from 1 to n)
For the states to be orthogonal
and for them to be normalised,
The phase velocity of ripples on a liquid surface is where S is the surface tension and ρ the density of the liquid. The group velocity of the ripples is given by :
As wave number,
We know that group velocity given by
The correct answer is:
The solutions of a free particle are :
V = 0 for a free particle
∴ Aeikx and Be–ikx are solution to the free particle
The correct answer is: Energy and momentum eigen functions with both positive and negative momentum eigenvalues.
The wave function for a particle constrained to move in 1D is shown in the graph below.
What is the probability that the particle would be found between x = 2 and x = 4 ?
The correct answer is: 13/16
The probability current density is given as Which of the following is true?
From the continuity equation
i.e. p is constant in time
Probability current density
The correct answer is: If then probability density is constant in time.
Smallest possible uncertainty in position of the electron moving with velocity 3 × 107 m/s. Given, h = 6.63 × 10–34 Js, m0 = 9.1 × 10–31 kg.
Given, v = 3 × 107 m/s
Let Δxmin be the minimum uncertainty in position of the electron and Δp the maximum uncertainty in the momentum of the electron.
Thus, we have,
= 0.03867 × 0.9949 × 10–10 m
= 3.8 × 10–12 m
The correct answer is: 3.8 × 10–12 m
Choose the correct statement
The correct answer is: All the energy eigenvalues are always real.
The wave function of the particle lies in which region?
The particle cannot exist outside the box, as it cannot have infinite amount of energy. Thus, it’s wave function is between 0 and L, where L is the length of the side of the box.
A linear harmonic oscillator of mass m oscillates with a frequency where k is its force constant. What is the minimum energy of the oscillator.
The energy of the linear harmonic oscillator is
This is a constant of motion. We can represent the constant value of E by means of averages of the kinetic and potential over a cycle of motion by writing.
The average value of x and p should vanish for an oscillating particle. So, we can identifywith the square of the corresponding uncertainties.
Since from the uncertainty principle To determine the minimum energy of the oscillator, we put
The minimum energy is
Choose the correct statement for a free particle with ψ(x) = Aeikx
V = 0 for a free particle
∴ Aeikx and Be–ikx (General solution)
∴ are solution to the free particle
The correct answer is: Energy eigenvalue is and momentum is
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