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Q : 11.3 The photoelectric cut-off voltage in a certain experiment is 1.5\hspace{1mm}V. What is the maximum kinetic energy of photoelectrons emitted?

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Since the photoelectric cut-off voltage is 1.5 V. The maximum Kinetic Energy (eV) of photoelectrons emitted would be 1.5 eV.

KEmax=1.5 eV

KEmac=2.4\times10-19 J

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