The photoelectric cut-off voltage in an experiment was found to be 1.5 V. The work function for the material used in the experiment was 4.2 eV. The maximum kinetic energy of the photoelectrons emitted was
Show Hint
- 1.5 eV
- 2.7 eV
- 4.2 eV
- 5.7 eV
The Correct Option is A
Solution and Explanation
The stopping potential \(V_0\) is related to maximum KE: \(KE_{max} = eV_0\).
Step 2: Detailed Explanation:
\(KE_{max} = eV_0 = e \times 1.5 \text{ V} = 1.5 \text{ eV}\)
Step 3: Final Answer:
Maximum KE of photoelectrons \(= \mathbf{1.5}\) eV.
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