Question

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength $\lambda_1$, then

• A. $\lambda \propto \dfrac{1}{\lambda_1}$
• B. $\lambda \propto \lambda_1$
• C. $\lambda \propto \lambda_1^2$
• D. $\lambda \propto \dfrac{1}{\lambda_1^2}$

Hint:

For negligible work function, photon energy directly equals kinetic energy of photoelectrons.

Answer 1

The Correct Option is C

Step 1: Use Einstein’s photoelectric equation.
Since work function is negligible, the entire photon energy converts into kinetic energy of photoelectrons:
\[ \frac{hc}{\lambda} = \frac{p^2}{2m} \]

Step 2: Express momentum using de-Broglie wavelength.
\[ p = \frac{h}{\lambda_1} \] \[ \Rightarrow \frac{hc}{\lambda} = \frac{h^2}{2m\lambda_1^2} \]

Step 3: Rearrangement.
\[ \lambda \propto \lambda_1^2 \]

Step 4: Conclusion.
The wavelength of the incident radiation is proportional to the square of the de-Broglie wavelength of the photoelectrons.