Question

A photon and an electron have equal energy $E$. $\lambda_{\mathrm{photon}} / \lambda_{\mathrm{electron}}$ is proportional to

• A. $\sqrt{E}$
• B. $\frac{1}{\sqrt{E}}$
• C. $\frac{1}{E}$
• D. does not depend upon $E$

Hint:

de Broglie wavelength for electron: $\lambda = \frac{h}{\sqrt{2mE}}$.

Answer 1

The Correct Option is B

To determine the proportionality of \(\frac{\lambda_{\mathrm{photon}}}{\lambda_{\mathrm{electron}}}\) with respect to energy \(E\), we need to analyze the wavelength expressions for both photons and electrons.

1. For a photon, the energy is related to its wavelength by the equation: \(E = \frac{hc}{\lambda_{\mathrm{photon}}}\) , where \(h\) is Planck's constant and \(c\) is the speed of light.
2. Rearranging gives the wavelength of the photon: \(\lambda_{\mathrm{photon}} = \frac{hc}{E}\).
3. For an electron, we use the de Broglie wavelength equation: \(\lambda_{\mathrm{electron}} = \frac{h}{p}\), where \(p\) is the momentum of the electron.
4. Since energy \(E\) is equal for both, the electron's energy is given by: \(E = \frac{p^2}{2m}\) (non-relativistic case).
5. Solving for momentum \(p\) gives: \(p = \sqrt{2mE}\), thus the de Broglie wavelength becomes: \(\lambda_{\mathrm{electron}} = \frac{h}{\sqrt{2mE}}\).
6. Now comparing \(\frac{\lambda_{\mathrm{photon}}}{\lambda_{\mathrm{electron}}}\): \(\frac{\lambda_{\mathrm{photon}}}{\lambda_{\mathrm{electron}}} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} = \frac{c\sqrt{2mE}}{E}\).
7. After simplification, it becomes: \(\frac{\lambda_{\mathrm{photon}}}{\lambda_{\mathrm{electron}}} \propto \frac{1}{\sqrt{E}}\).

Hence, the correct answer is that \(\frac{\lambda_{\mathrm{photon}}}{\lambda_{\mathrm{electron}}}\) is proportional to \(\frac{1}{\sqrt{E}}\).