the de broglie wavelength of an electron and proton are same which quantity will be same for both

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If an electron and a proton have the same de

Click here👆to get an answer to your question ✍️ If an electron and a proton have the same de - Broglie wavelength, then the kinetic energy of the electron is :

Question

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is :

Zero

Less than that of a proton

More than that of a proton

Equal to that of a proton

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Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

de Broglie wavelength λ=

2mK h We get K= 2mλ 2 h 2 ⟹ K∝ m 1 We know that m p >m e ∴ K p K e = m e m p >1 ⟹ K e >K p

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The de Broglie wavelength of electron and proton are same. Which quantity will be same for both ?

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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

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. Ans: Hint To answer this ...

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

(D) a and b both are correct

Last updated date: 17th Mar 2023

• Total views: 262.2k • Views today: 7.42k Answer Verified 262.2k+ views 2 likes

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

⇒E=hν →1 ⇒E=hν →1

This is taken from Planck's energy equation.

Where,

E is the energy of the photon

H is the Planck constant

ν ν

is the frequency of the photon

And ⇒E=m c 2 →2 ⇒E=mc2 →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

⇒hν=m c 2 →3 ⇒hν=mc2 →3

We know that frequency is given by

⇒ν= c λ ⇒ν=cλ

Substituting this in equation 3

⇒h c λ =m c 2 ⇒hcλ=mc2 ⇒λ= h mc ⇒λ=hmc

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

⇒λ= h mv ⇒λ=hmv or ⇒λ= h p →4 ⇒λ=hp →4 ⇒p=mv ⇒p=mv λ λ

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.

⇒λ= h p ⇒λ=hp

Rearranging equation, we get

⇒p= h λ ⇒p=hλ

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.

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