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
Answer & explanation
Correct answer: option 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:
$\frac{\lambda_1}{\lambda_2} = \frac{ \frac{h}{\sqrt{2mE}} }{ \frac{hc}{E} } = \frac{E}{c\sqrt{2mE}} = \frac{\sqrt{E}}{c\sqrt{2m}}$
∴ $\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$