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

Photoelectric cut-off voltage, V₀ = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as:

Where,

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

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10⁻¹⁹ J.