Question

the photoelectric cutoff of the voltage in a circuit experiment is 1.5 volt what is the maximum kinetic energy of photoelectrons emitted ​

Answer 1

Answer:

2.4 × 10-¹⁹J

Explanation:

Photoelectric cut-off voltage, V0 = 1.5 V

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

Where,

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

∴ Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J

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

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