Question

The wavelength \(\lambda _{e}\) of an electron and \(\lambda _{P}\) of a photon of same energy E are related as

• A. \(\lambda _{P}\ \alpha \ \sqrt{\lambda _{e}}\)
• B. \(\lambda _{P}\ \alpha \ \frac{1}{\lambda _{e}}\)
• C. \(\lambda _{P}\ \alpha \ \lambda _{e}^{2}\)
• D. \(\lambda _{P}\ \alpha \ \lambda _{e}\)

Hint:

Write each wavelength in terms of the common energy E, then eliminate E.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A photon and an electron have the same energy \(E\). We write the wavelength of each in terms of \(E\).

Step 2: Photon
For a photon, \(E=\dfrac{hc}{\lambda_P}\), so \(\lambda_P=\dfrac{hc}{E}\).

Step 3: Electron
For an electron, \(E=\dfrac{p^{2}}{2m}\) and \(\lambda_e=\dfrac{h}{p}\). So \(E=\dfrac{h^{2}}{2m\lambda_e^{2}}\).

Step 4: Eliminate E
\(\lambda_P=\dfrac{hc}{E}=hc\times\dfrac{2m\lambda_e^{2}}{h^{2}}=\dfrac{2mc}{h}\,\lambda_e^{2}\). So \(\lambda_P\propto\lambda_e^{2}\).

Final Answer:
The photon wavelength is proportional to the square of the electron wavelength. \[ \boxed{\text{(C) }\lambda_P\ \propto\ \lambda_e^{2}} \]