Question

When a certain metal surface is illuminated with a light of wavelength \(\lambda\), the stopping potential is \(V\). When the same surface is illuminated by light of wavelength \(2\lambda\), the stopping potential is \(\left(\frac{V}{3}\right)\). The threshold wavelength for the surface is

• A. \(\frac{8\lambda }{3}\)
• B. \(\frac{4\lambda }{3}\)
• C. \(4\lambda\)
• D. \(6\lambda\)

Hint:

Write Einstein's equation for both wavelengths, subtract to eliminate the work function, then find the threshold wavelength.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Einstein's equation gives \(eV_s=\frac{hc}{\lambda}-\phi\), where \(\phi=\frac{hc}{\lambda_0}\) is the work function.

Step 2: Write both cases
For \(\lambda\): \(eV=\frac{hc}{\lambda}-\phi\).
For \(2\lambda\): \(\frac{eV}{3}=\frac{hc}{2\lambda}-\phi\).

Step 3: Solve
Subtracting: \(\frac{2eV}{3}=\frac{hc}{2\lambda}\), so \(eV=\frac{3hc}{4\lambda}\).
Then \(\phi=\frac{hc}{\lambda}-\frac{3hc}{4\lambda}=\frac{hc}{4\lambda}\).

Step 4: Threshold wavelength
Since \(\phi=\frac{hc}{\lambda_0}\), we get \(\lambda_0=4\lambda\), option (C).

Final Answer:
The threshold wavelength is 4 times lambda. \[ \boxed{\text{(C) }4\lambda} \]