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.

 

 

Answers (1)

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}

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