Question:
The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?
Solution:
Photoelectric cut-off voltage, Vo = 1.5 V
For emitted photoelectrons, the maximum kinetic energy is:
Ke = eVo
Where,
e = charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J
Therefore, 2.4 x 10-19 J is the maximum kinetic energy emitted by the photoelectrons.
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