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: Understanding the Concept
Einstein's equation: \(\dfrac{hc}{\lambda}=\dfrac{hc}{\lambda_{0}}+K\). The kinetic energy is \(K=\dfrac{p^{2}}{2m}\).

Step 2: Case 1 and Case 2
Case 1: \(K=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_{0}}\right)\).
When the momentum becomes \(2p\), the kinetic energy becomes \(4K\): \(4K=hc\left(\dfrac{1}{\lambda'}-\dfrac{1}{\lambda_{0}}\right)\).

Step 3: Calculation
\(\dfrac{1}{\lambda'}-\dfrac{1}{\lambda_{0}}=4\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_{0}}\right)\)
\(\dfrac{1}{\lambda'}=\dfrac{4}{\lambda}-\dfrac{3}{\lambda_{0}}=\dfrac{4\lambda_{0}-3\lambda}{\lambda\lambda_{0}}\)

Step 4: Result
\(\lambda'=\dfrac{\lambda\lambda_{0}}{4\lambda_{0}-3\lambda}\)

Final Answer:
The required wavelength is lambda times lambda0 divided by (4 lambda0 minus 3 lambda). \[ \boxed{\text{(A) }\dfrac{\lambda\lambda_{0}}{4\lambda_{0}-3\lambda}} \]