The de-Broglie wavelength of a particle moving with a velocity 2.25 x 10⁸ m/s is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of the photon is: (Velocity of light = 3 × 10⁸ m/s)

$\lambda =\frac{h}{mv}$  is the de Broglie wavelength. 

The energy of a photon of this wavelength is

$\mathrm{E}=\mathrm{h}\nu$$=\frac{hc}{\lambda }=\frac{hc}{h/mv}=mvc$             …………….(i)

Kinetic energy of the particle is

$E\text{'}=\frac{1}{2}m{v}^2$                                  ………………..(ii)

From (i) and (ii), we have

$$\begin{gathered} \frac{E\text{'}}{E}=\frac{\frac{1}{2}m{v}^2}{mvc} \\ =\frac{v}{2c}=\frac{2.25\times {10}^8}{2\times \left(3\times {10}^8\right)}=\frac{2.25}{6}=\frac{3}{8} \end{gathered}$$