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: Use a short notation
Let $x=\dfrac{1}{\lambda}$ and $y=\dfrac{1}{\lambda_0}$. Then $V\propto x-y$ and $\dfrac{V}{8}\propto \dfrac{x}{4}-y$.

Step 2: Form the equation
$x-y=8\left(\dfrac{x}{4}-y\right)=2x-8y$.

Step 3: Solve
$7y=x$, so $\dfrac{7}{\lambda_0}=\dfrac{1}{\lambda}$ and $\lambda_0=7\lambda$. This is option (D).

Final Answer:
Solving the two equations gives a threshold wavelength of 7 lambda. \[ \boxed{\text{(D) }7\lambda} \]