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: Use energy form
Let $E=\frac{hc}{\lambda}$. Then $eV=E-\phi$ and $\frac{eV}{3}=\frac{E}{2}-\phi$.

Step 2: Eliminate eV
Multiplying the second by 3: $eV=\frac{3E}{2}-3\phi$.
Equating: $E-\phi=\frac{3E}{2}-3\phi$, so $2\phi=\frac{E}{2}$ and $\phi=\frac{E}{4}$.

Step 3: Result
$\frac{hc}{\lambda_0}=\frac{1}{4}\frac{hc}{\lambda}$, so $\lambda_0=4\lambda$. Option (C) is correct.

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