Question

A radiation of wave length 2500 A° is incident on a metal plate whose work function is 3.5eV. Then the potential required to stop the fastest photo electrons emitted by the surface is(h = 6.63 × 10^-34 Js and c = 3 × 10^8 m/s)

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Answer 1

Explanation:

Energy of photon E=hν=

λ

hc

Given:

λ=2500×10

−10

m

E=

2500×10

−10

6.63×3×10

−26

J

E=

1.6×10

−19

0.007956×10

−16

eV=4.97eV

Work function of metal is 3.5eV.

So maximum kinetic energy of the photoelectron after emission K=E−W=(4.97−3.5)eV=1.47eV

So the voltage required to stop the fastest photoelectron is 1.47V.

Answer 2

Answer:

check the picture sissy

thank you for helping me sisy

third option is correct