An electron and a photon have same wavelength lambda, what is the ratio of their kinetic energies ??

2 Answers
Nov 1, 2017

Check for error below. See the other solution.

Explanation:

We know that de-Broglie wavelength lambda of a particle of mass m and momentum p is given by the expression

lambda=h/p
where h is Planck's Constant.

Kinetic energy is given as

KE=p^2/(2m)

In terms of de-Broglie wavelength lambda

KE=(h/lambda)^2/(2m)
=>KE=h^2/(2mlambda^2)

Ratio of kinetic energies of electron and proton

(KE_e)/(KE_p)=(h^2/(2mlambda^2))_e/(h^2/(2mlambda^2))_p

When both have same wavelength above reduces to

(KE_e)/(KE_p)=m_p/m_e

Jul 31, 2018

I misread sf"photon" as proton. Correct answer is as below.

Explanation:

We know that de-Broglie wavelength lambda of a particle of mass m and momentum p is given by the expression

lambda=h/p
where h is Planck's Constant.

Kinetic is given as

KE=p^2/(2m)

In terms of de-Broglie wavelength lambda

KE=(h/lambda)^2/(2m)
=>KE=h^2/(2mlambda^2)

We know that photon is a mass less particle. Its Kinetic Energy is same as its Energy which is given by the expression

E=(hc)/lambda
where c is velocity of light

Ratio of kinetic energies of electron and photon

(KE_e)/(KE_p)=(h^2/(2m_elambda_e^2))/((hc)/lambda_p)

When both have same wavelength lambda above reduces to

(KE_e)/(KE_p)=h/(2m_elambdac)