Question

Find the maximum kinetic
energy of photo electrons
ejected from a metal surface
when a light of frequency
1x10^15Hz is incident on it.(
given that the threshold
frequency of metal is
0.66x10^15Hz) *
O zev
3 ev
oo
2.83eV
1.38eV​

Answer 1

Answer:

◆ Answer -

ν0 = 10^15 Hz

● Explaination -

# Given -

KE = 6.63×10^-19 J

ν = 2×10^15 Hz

# Solution -

Work done by the photoelectric process is given by -

W = Φ0 + KE

hν = hν0 + KE

hν0 = hν - KE

ν0 = ν - KE/h

Now substitute values,

ν0 = 2×10^15 - 6.63×10^-19 / 6.63×10^-34

ν0 = 2×10^15 - 10^15

ν0 = 10^15 Hz

Therefore, threshold frequency of the metal is 10^15 Hz.

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