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Wave Optics

Question

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

Solution

Given, cut-off voltage, V_o = 1.5 V

Maximum kinetic energy of photoelectrons emitted is,

K . E ._{\max} = {eV}_{0} = 1.5 \times 1.6 \times 10^{- 19} J = 2.4 \times 10^{- 19} J

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