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Question
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
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Solution
Photoelectric cut-off voltage, V0 = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as:
Ke = eV0
Where,
e = Charge on an electron = 1.6 × 10−19 C
∴ Ke = 1.6 × 10−19 × 1.5
= 2.4 × 10−19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10−19 J.
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