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 ______. ($h = 6.62 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C and } c = 3 \times 10^8 \text{ m/s}$)
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 ______. ($h = 6.62 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C and } c = 3 \times 10^8 \text{ m/s}$)
Updated On: Apr 12, 2026
- 3.68
- 4.68
- 5.68
- 2.68
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
Using Einstein's photoelectric equation, we need to calculate the energy of incident photons, the kinetic energy of emitted electrons (from stopping potential), and find the work function.
Step 2: Key Formula or Approach:
1. Energy of photon $E = \frac{hc}{\lambda}$.
2. Max kinetic energy $K_{max} = eV_{stopping}$.
3. Work function $\Phi = E - K_{max}$.
Step 3: Detailed Explanation:
Energy of incident photon ($E$):
\[ E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}} = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 0.06 \times 10^{-17} = 6 \times 10^{-19} \text{ J} \]
Maximum kinetic energy ($K_{max}$):
\[ K_{max} = e \times 0.2 \text{ V} = 0.2 \times (1.6 \times 10^{-19} \text{ C}) = 0.32 \times 10^{-19} \text{ J} \]
Work function ($\Phi$):
\[ \Phi = E - K_{max} = (6 \times 10^{-19}) - (0.32 \times 10^{-19}) = 5.68 \times 10^{-19} \text{ J} \]
Thus, $\alpha = 5.68$.
Step 4: Final Answer:
The value of $\alpha$ is 5.68.
Using Einstein's photoelectric equation, we need to calculate the energy of incident photons, the kinetic energy of emitted electrons (from stopping potential), and find the work function.
Step 2: Key Formula or Approach:
1. Energy of photon $E = \frac{hc}{\lambda}$.
2. Max kinetic energy $K_{max} = eV_{stopping}$.
3. Work function $\Phi = E - K_{max}$.
Step 3: Detailed Explanation:
Energy of incident photon ($E$):
\[ E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}} = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 0.06 \times 10^{-17} = 6 \times 10^{-19} \text{ J} \]
Maximum kinetic energy ($K_{max}$):
\[ K_{max} = e \times 0.2 \text{ V} = 0.2 \times (1.6 \times 10^{-19} \text{ C}) = 0.32 \times 10^{-19} \text{ J} \]
Work function ($\Phi$):
\[ \Phi = E - K_{max} = (6 \times 10^{-19}) - (0.32 \times 10^{-19}) = 5.68 \times 10^{-19} \text{ J} \]
Thus, $\alpha = 5.68$.
Step 4: Final Answer:
The value of $\alpha$ is 5.68.
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