Question

The wavelength $\lambda_e$ of an electron and $\lambda_P$ of a photon of same energy E are related by:

• A. $ \lambda_P \propto \frac{1}{\sqrt{\lambda_e}}$
• B. $ \lambda_P \propto \lambda^2_e$
• C. $ \lambda_P \propto \lambda_e$
• D. $ \lambda_P \propto \sqrt{ \lambda_e}$

Answer 1

The Correct Option is B

 To understand the relationship between the wavelength of an electron \(\lambda_e\) and the wavelength of a photon \(\lambda_P\) when both have the same energy E, we can start by considering the expressions for the wavelengths in terms of their respective properties.

Step-by-Step Explanation:

1. Wavelength of an Electron:
   • From De Broglie's hypothesis, the wavelength of an electron is given by: \(\lambda_e = \frac{h}{p_e}\), where h is the Planck's constant, and p_e is the momentum of the electron.
   • Momentum p_e is related to the kinetic energy of the electron by: \(E = \frac{p_e^2}{2m}\), leading to \(p_e = \sqrt{2mE}\).
   • Substituting back, we get: \(\lambda_e = \frac{h}{\sqrt{2mE}}\).
2. Wavelength of a Photon:
   • The wavelength of a photon is given by: \(\lambda_P = \frac{hc}{E}\), where c is the speed of light.
3. Relating the Two Wavelengths:
   • We have: \(\lambda_e = \frac{h}{\sqrt{2mE}}\) and \(\lambda_P = \frac{hc}{E}\).
   • Substitute E from the photon wavelength into the equation for the electron: \(E = \frac{hc}{\lambda_P}\) gives \(\lambda_e = \frac{\lambda_P \sqrt{2m \lambda_P}}{c}\).
   • This yields a relationship between \(\lambda_e\) and \(\lambda_P\) as: \(\lambda_P \propto \lambda_e^2\).

The correct answer is: \(\lambda_P \propto \lambda_e^2\), as deduced from the above analysis.

Other options do not satisfy this derived proportionality relation based on the equations involved for photon and electron wavelengths given the same energy.