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Question -

Thephotoelectric cut-off voltage in a certain experiment is 1.5 V. What is themaximum kinetic energy of photoelectrons emitted?



Answer -

Photoelectric cut-off voltage, V0 =1.5 V

Themaximum kinetic energy of the emitted photoelectrons is given as:

Where,

e = Charge on an electron = 1.6 × 10−19 C

Therefore,the maximum kinetic energy of the photoelectrons emitted in the givenexperiment is 2.4 × 10−19 J.

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