Question

${\underline{\underline{\tt{QUESTION}}}}:-$
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?​

Answer 1

Answer:

Kinetic energy = 2.4 × 10^-19 J

Explanation:

The magnitude of cut off voltage in V is equal to magnitude of kinetic energy in eV.

•°• KE = V'

V' = cut-off voltage = 1.5 V

KE = 1.5 eV = 1.5 × 1.6 × 10^-19 J [ e = 1.6 × 10^-19 ]

KE = 2.4 × 10^-19 J

Answer 2

Answer:

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