The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let $λ_1$ be the de- Broglie wavelength of the proton and $λ_2$ be the wavelength of the photon. The ratio $λ_1/λ_2$ is proportional to the following |
$E^0$ $E^{1/2}$ $E^{-2}$ $E^{-1}$ |
$E^{1/2}$ |
The correct answer is Option (2) → $E^{1/2}$ Given photon energy: $E = h\nu = \frac{hc}{\lambda_2}$ For the proton, kinetic energy $K = E = \frac{1}{2}mv^2$ Its de-Broglie wavelength: $\lambda_1 = \frac{h}{mv}$ From kinetic energy: $v = \sqrt{\frac{2E}{m}}$ Substitute into $\lambda_1$: $\lambda_1 = \frac{h}{m\sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2mE}}$ Now, the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{ \frac{h}{\sqrt{2mE}} }{ \frac{hc}{E} } = \frac{E}{c\sqrt{2mE}} = \frac{\sqrt{E}}{c\sqrt{2m}}$ ∴ $\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$ |