# A photon and a proton have the same de-Broglie wavelength $\lambda$. Prove that the energy of the photon is $(2m\lambda c/h)$ times the kinetic energy of the proton.

We know that the energy of a photon is given by;

$E_{P}=\frac{hc}{\lambda }$

Where, $c$ is the speed of light

$h$ is plank's constant and

$\lambda$ is the wavelength

Such that, for proton, the wavelength will be,

$\lambda =\frac{h}{mv}$

Or

$mv=\frac{h}{\lambda }$

Now, the expression for the kinetic energy of the proton

$E_{K}=\frac{1}{2}mv^{2}$ ______(1)

On placing the value of $mv$ in equation, (1), we get

$E_{K}=\frac{1}{2}\frac{h^{2}}{m\lambda ^{2}}$

Substituting

$\frac{h}{\lambda}=\frac{E_p}{C}$

Then

$E_{K}=\frac{1}{2}\frac{h}{m\lambda}\times\frac{E_p}{C}$

or

$E_{P}=\left [ \frac{2m\lambda c}{h} \right ]E_{K}$

## Related Chapters

### Preparation Products

##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
##### Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
##### Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-