Question

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is \(3.2\ \text{V}\). If a second light having wavelength twice of the first light is used, the stopping potential drops to \(0.7\ \text{V}\). The wavelength of the first light is ________ m.

• A. \(2.2 \times 10^{-8}\)
• B. \(3.1 \times 10^{-7}\)
• C. \(2.5 \times 10^{-7}\)
• D. \(2.9 \times 10^{-8}\)

Hint:

Stopping potential depends on frequency, not on intensity of incident light.

Answer 1

The Correct Option is C

Step 1: Write Einstein’s photoelectric equation.
\[ eV_s = \frac{hc}{\lambda} - \phi \]
Step 2: Write equations for both cases.
For first light,
\[ e(3.2) = \frac{hc}{\lambda} - \phi \] For second light with wavelength \(2\lambda\),
\[ e(0.7) = \frac{hc}{2\lambda} - \phi \]
Step 3: Subtract the two equations.
\[ e(3.2 - 0.7) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \] \[ e(2.5) = \frac{hc}{2\lambda} \]
Step 4: Substitute constants and calculate wavelength.
\[ \lambda = \frac{hc}{2e(2.5)} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 1.6 \times 10^{-19} \times 2.5} \] \[ \lambda \approx 2.5 \times 10^{-7}\ \text{m} \]