Question:

A photon and an electron have an equal energy ' \(\text{E}\) '. The ratio of wavelength ' \(\lambda_\text{p}\) ' of photon to that of electron ' \(\lambda_\text{e}\) ' is proportional to

Show Hint

Photon: $\lambda \propto \frac{1}{E}$ Electron: $\lambda \propto \frac{1}{\sqrt{E}}$
Updated On: May 8, 2026
  • \(\sqrt{\text{E}}\)
  • \(\frac{1}{\sqrt{\text{E}}}\)
  • \(\frac{1}{\text{E}}\)
  • \(\text{E}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation


Concept: For photon: \[ E = \frac{hc}{\lambda_p} \Rightarrow \lambda_p \propto \frac{1}{E} \] For electron: \[ E = \frac{p^2}{2m} \Rightarrow p \propto \sqrt{E} \] \[ \lambda_e = \frac{h}{p} \propto \frac{1}{\sqrt{E}} \]

Step 1:
Take ratio. \[ \frac{\lambda_p}{\lambda_e} \propto \frac{1/E}{1/\sqrt{E}} = \sqrt{E} \] Final Answer: Option (A)
Was this answer helpful?
0
0

Top MHT CET Dual nature of radiation and matter Questions

View More Questions