Question

The energy of a photo is equal to the kinetic energy of proton. Let ${\lambda }_1$ be the de-Broglie wavelength of the photon and ${\lambda }_2$ be the wavelength of the radiation and the de-Broglie wavelength of a photon of that radiation?

Answer 1

We know that,

The de- Broglie wave length is

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

Now, photon energy

$E=h\nu$

So, the de- Broglie wave length of photon is

${\lambda }_1=\frac{h}{p}$

${\lambda }_1=\frac{h}{\sqrt{2mE}}....\left(I\right)$

Now, the de- Broglie wave length of radiation is

$E=h{\nu }_2$

$E=\frac{hc}{{\lambda }_2}$

${\lambda }_2=\frac{hc}{E}....\left(II\right)$

Now, from equation (I) and (II)

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}$

$\frac{{\lambda }_1}{{\lambda }_2}\sim \frac{E}{\sqrt{E}}$

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{E}$

Hence, the de-Broglie wavelength of a photon of that radiation is $\sqrt{E}$