Question

A photoemissive substance is illuminated with a radiation of wavelength \(λ_i\) so that it releases electrons with de-Broglie wavelength \(λ_e\). The longest wavelength of radiation that can emit photoelectron is \(λ_0\). Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)

• A. \((\text{h}λ_i/2\text{mc})^{\frac{1}{2}}\)
• B. \((\text{h}λ_0/2\text{mc})^{\frac{1}{2}}\)
• C. \([\text{h}/2\text{mc}(\frac{1}{λ_i}-\frac{1}{λ_0})]^{\frac{1}{2}}\)
• D. \([\text{h}/[2\text{mc}(\frac{1}{λ_i}-\frac{1}{λ_0})]^{\frac{1}{2}}\)

Hint:

Kinetic energy is hc(1/lambda_i minus 1/lambda_0); then use lambda = h over root(2mK).

Answer 1

The Correct Option is C

Step 1: Momentum route
$p = \sqrt{2mK}$, so $\lambda_e^2 = \frac{h^2}{2mK}$.

Step 2: Insert K
$\lambda_e^2 = \frac{h^2}{2mhc(1/\lambda_i-1/\lambda_0)} = \frac{h}{2mc(1/\lambda_i-1/\lambda_0)}$. Taking the square root gives option (C).

Final Answer:
Option C. \[ \boxed{\text{(C)}\ \left[\frac{h}{2mc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)}\right]^{1/2}} \]