Question

When a metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is \(V\). If the surface is now illuminated with a radiation of wavelength \(4\lambda\), the stopping potential \(\left(\frac{V}{8}\right)\). The threshold wavelength for the surface is

• A. \(\lambda\)
• B. \(3\lambda\)
• C. \(5\lambda\)
• D. \(7\lambda\)

Hint:

Write Einstein's equation for both wavelengths and subtract to eliminate the work function.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Einstein's equation gives \(eV_s=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}\), where \(\lambda_0\) is the threshold wavelength.

Step 2: Write both cases
For wavelength \(\lambda\): \(eV=hc\left(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\right)\).
For wavelength \(4\lambda\): \(\dfrac{eV}{8}=hc\left(\dfrac{1}{4\lambda}-\dfrac{1}{\lambda_0}\right)\).

Step 3: Eliminate hc and solve
Divide the first equation by the second: \(8=\dfrac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{4\lambda}-\frac{1}{\lambda_0}}\). So \(\dfrac{8}{4\lambda}-\dfrac{8}{\lambda_0}=\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}\).

Step 4: Find the threshold wavelength
\(\dfrac{2}{\lambda}-\dfrac{1}{\lambda}=\dfrac{7}{\lambda_0}\Rightarrow \lambda_0=7\lambda\), option (D).

Final Answer:
The threshold wavelength is 7 times lambda. \[ \boxed{\text{(D) }7\lambda} \]