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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Given :
Photo electric cut off voltage, Vo= 1.5V
Formula to be used :
The maximum kinetic energy of photo electrons emitted:
Ke= eVo
where
e= charge of an electron = 1.6 x 10 ⁻¹⁹ c
Ke=1.6x 10⁻¹⁹ x 1.5
=2.4 x 10⁻¹⁹ J
∴ the maximum kinetic energy of photo electrons emitted in the experiment is 2.4 x 10⁻¹⁹ J
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