Question

A light of wavelength \(\lambda\) \((\lambda \lt \lambda _{0})\) falls on a metal surface. The momentum of the fastest moving electron is \(p\). In order to have the momentum of an electron equal to \(2p\), the required wavelength \(\lambda '\) of incident light must be

• A. \(\frac{\lambda \lambda _{0}}{4\lambda _{0}-3\lambda }\)
• B. \(\frac{4\lambda _{0}-3\lambda }{\lambda \lambda _{0}}\)
• C. \(\frac{\lambda _{0}}{4\lambda _{0}-3\lambda }\)
• D. \(\frac{4\lambda _{0}-3\lambda }{\lambda }\)

Hint:

Kinetic energy is proportional to the square of momentum, so doubling the momentum makes the kinetic energy four times larger.

Answer 1

The Correct Option is A

Step 1: Formula-first approach
$p\propto\sqrt{K}$, so $p\to 2p$ means $K\to 4K$.

Step 2: Use the work function
The work function is $\dfrac{hc}{\lambda_{0}}$. So $4\left(\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_{0}}\right)=\dfrac{hc}{\lambda'}-\dfrac{hc}{\lambda_{0}}$.

Step 3: Compare the options
Solving gives $\lambda'=\dfrac{\lambda\lambda_{0}}{4\lambda_{0}-3\lambda}$.
Options B and D have the wrong dimensions for a wavelength, and option C is missing $\lambda$ in the numerator.

Final Answer:
Option A is the only expression with the dimensions of a wavelength that satisfies the energy balance. \[ \boxed{\text{(A) }\dfrac{\lambda\lambda_{0}}{4\lambda_{0}-3\lambda}} \]