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: Kinetic energy
Work function \(\phi = \frac{hc}{\lambda_0}\). So \(K = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)\).

Step 2: de Broglie wavelength
\(\lambda_e = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mhc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)}}\).

Step 3: Simplify
\[ \lambda_e = \left[\frac{h}{2mc\left(\frac1{\lambda_i}-\frac1{\lambda_0}\right)}\right]^{1/2} \]
This matches option (C).

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