Question

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is $\alpha \times 10^{-19} \text{ J}$. The value of $\alpha$ is ______. ($h = 6.62 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C and } c = 3 \times 10^8 \text{ m/s}$)

• A. 3.68
• B. 4.68
• C. 5.68
• D. 2.68

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The topic of this question is the Photoelectric Effect.
Using Einstein's photoelectric equation, we are to determine the work function of a metal given the incident light wavelength and the stopping potential.
Step 2: Key Formula or Approach:
1. Energy of the photon: $E = \frac{hc}{\lambda}$.
2. Maximum kinetic energy: $K_{max} = eV_{stopping}$.
3. Work function: $\Phi = E - K_{max}$.
Step 3: Detailed Explanation:
First, find the energy of the incident photon in Joules:
\[ E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}} = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 6 \times 10^{-19} \text{ J} \]
Next, find the maximum kinetic energy using the stopping potential:
\[ K_{max} = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19} \text{ J} \]
Finally, solve for the work function ($\Phi$):
\[ \Phi = E - K_{max} = (6 \times 10^{-19}) - (0.32 \times 10^{-19}) = 5.68 \times 10^{-19} \text{ J} \]
Hence, the multiplier $\alpha = 5.68$.
Step 4: Final Answer:
The value of $\alpha$ is 5.68.