Question

In a photoelectric experiment, the stopping potential for incident light of wavelength 4000 Å is 2 V. If the wavelength is changed to 3000 Å, the new stopping potential will be approximately:

• A. 2 V
• B. 3.03 V
• C. 4.14 V
• D. 1.5 V

Hint:

Shorter wavelength means higher energy photons, which results in higher stopping potential.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to Einstein's photoelectric equation, the energy of an incident photon equals the sum of the work function and the maximum kinetic energy of the emitted electron. Maximum kinetic energy is equivalent to \( e \times V_{stopping} \).
Step 2: Key Formula or Approach:
1. \( \frac{hc}{\lambda} = \phi + eV_s \).
2. Photon energy in eV \(\approx \frac{12420}{\lambda(\text{\AA})}\) (or \( \frac{12400}{\lambda} \)).
Step 3: Detailed Explanation:
Calculate energy for 4000 \AA: \( E_1 = \frac{12420}{4000} \approx 3.105 \text{ eV} \). Using \( E_1 = \phi + eV_{s1} \): \( 3.105 = \phi + 2 \implies \phi = 1.105 \text{ eV} \). Now calculate energy for 3000 \AA: \( E_2 = \frac{12420}{3000} = 4.14 \text{ eV} \). Using \( E_2 = \phi + eV_{s2} \): \( 4.14 = 1.105 + V_{s2} \) \( V_{s2} = 4.14 - 1.105 = 3.035 \text{ V} \).
Step 4: Final Answer:
The new stopping potential is approximately 3.03 V.