Question

If a proton and electron have same de broglie wavelength, then
(A) Momentum of electron = momentum of proton
(B) Velocity of electron > velocity of proton
(C) Momentum of electron > momentum of proton
(D) a and b both are correct

Answer 1

Hint
To answer this question, we should have known about the de broglie's wave equation. Then we can equate the wavelength of the proton and electron and say which of the given statements would satisfy the condition.

Complete step by step answer
Louis de Broglie, a French Physicist, in 1924, proposed the idea that like photons, all material particles such as electron, proton, atom, molecule, a piece of chalk, a piece of stone or iron ball possessed both wave character as well as particle character. He said that the wave associated with a particle is called a matter wave.
The wavelength of the wave associated with any material particle was calculated by analogy with photon
If we take a photon, and assume it to have wave character, then the energy of photon is given by
$ \Rightarrow E = h\nu {\text{ }} \to {\text{1}}$
This is taken from Planck's energy equation.
Where,
E is the energy of the photon
H is the Planck constant
$\nu $ is the frequency of the photon
And
$ \Rightarrow E = m{c^2}{\text{ }} \to {\text{2}}$
This is taken from the Einstein’s energy equation
Where,
m is the mass of photon
c is the velocity of light.
Equating the equation 1 and 2 we get
$ \Rightarrow h\nu = m{c^2}{\text{ }} \to {\text{3}}$
We know that frequency is given by
$ \Rightarrow \nu = \dfrac{c}{\lambda }$
Substituting this in equation 3
$ \Rightarrow h\dfrac{c}{\lambda } = m{c^2}$
$ \Rightarrow \lambda = \dfrac{h}{{mc}}$
This equation gives the wavelength of the photon.
We took the particle as a photon for an example. But this equation is applicable for every particle.
De Broglie said that the above equation is applicable to any material particle. The mass of the photon is replaced by the mass of the material particle and the velocity “c” of the photon is replaced by the velocity v of the material particle. Thus, for any material particle, the de Broglie’s equation is given by
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$
or
 $ \Rightarrow \lambda = \dfrac{h}{p}{\text{ }} \to {\text{4}}$
$\Rightarrow p = mv$
$\lambda $ is the wavelength of the particle
h is the Planck’s constant
m is the mass
V is the velocity
P is the momentum
Now in the question it is stated that a proton and electron have the same de broglie wavelength.
$ \Rightarrow \lambda = \dfrac{h}{p}{\text{ }}$
Rearranging equation, we get
$ \Rightarrow p = \dfrac{h}{\lambda }{\text{ }}$
So if two particles have same wavelength then their velocity and momentum will be equal
If a proton and electron have the same de broglie wavelength their momentum will be equal.
Hence the correct answer is option (A) Momentum of electron = momentum of proton.

Note
The de broglie's wave equation is only applicable for micro particles. It is not applicable for big particles like cricket balls, because the wavelength of the wave associated with them is too small to be calculated.