Question

When a photosensitive material is illuminated by photons of energy 3.1 eV, the stopping potential of the photoelectrons is 1.7 V. When the same photosensitive material is illuminated by photons of energy 2.5 eV, the stopping potential of the photoelectrons is

• A. 1.8 V
• B. 1.4 V
• C. 1.1 V
• D. 1.3 V

Hint:

If energies are given in eV and stopping potential in V, you can treat them as numerically equal in the equation $E = \phi + V_s$.

Answer 1

The Correct Option is C

Step 1: Einstein's Photoelectric Equation:
$E = \phi + K_{max} = \phi + e V_s$ where $E$ is photon energy, $\phi$ is work function, $V_s$ is stopping potential.
Step 2: Find Work Function ($\phi$):
From the first case: $3.1 \text{ eV} = \phi + e(1.7 \text{ V})$. Since $e(1 \text{ V}) = 1 \text{ eV}$, we can write: $3.1 = \phi + 1.7$ $\phi = 3.1 - 1.7 = 1.4 \text{ eV}$.
Step 3: Calculate New Stopping Potential:
From the second case: $E' = 2.5 \text{ eV}$. $2.5 = \phi + e V_s'$. $2.5 = 1.4 + e V_s'$. $e V_s' = 1.1 \text{ eV}$. $V_s' = 1.1 \text{ V}$.