Question 3

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Answer

Photoelectric cut-off voltage, V_{0} = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as: *K _{e} = eV_{o}*

Where,

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

∴ K_{e} = 1.6 x 10^{-19} x 1.5 = 2.4 x 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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Ambily
2019-01-28 22:00:48

Thanq you so much

- NCERT Chapter