Question

When a photosensitive metal surface is illuminated with radiation of wavelength ' $\lambda_1$ ', the stopping potential is ' $V_1$ '. If the same surface is illuminated with radiation of wavelength ' $3\lambda_1$ ', the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is

• A. $\frac{3}{2} \lambda_1$
• B. $2\lambda_1$
• C. $5\lambda_1$
• D. $6\lambda_1$

Hint:

For photoelectric effect: \[ eV_s=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \] If two stopping potentials are given, convert the relation into a simple equation using reciprocals.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Einstein's photoelectric equation relates incident energy to work function and stopping potential energy.
Step 2: Key Formula or Approach:
$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV_s$.
Step 3: Detailed Explanation:
1. $hc/\lambda_1 = hc/\lambda_0 + eV_1$.
2. $hc/(3\lambda_1) = hc/\lambda_0 + eV_1/6$.
Multiply (2) by 6: $2hc/\lambda_1 = 6hc/\lambda_0 + eV_1$.
Subtract (1) from this: $hc/\lambda_1 = 5hc/\lambda_0 \implies \lambda_0 = 5\lambda_1$.
Step 4: Final Answer:
The threshold wavelength is $5\lambda_1$.