Light of wavelength a and power P is incident on a caesium surface having work function w.If 0.5% of
the incident photons produce photoelectrons, then the photoelectric current will be
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Answer:
E = hν
=
λ
hc
=
4560×10
−10
J
6.6×10
−34
×3×10
8
= 4.34×10
−19
J
1mW of light energy is equivalent to
(4.34×10
−19
)
10
−3
= 2.3×10
15
phtons/s
The quantum efficiency is 0.5%
The means that only 0.5% of these photons release photoelectrons.
so, the number of electrons released form the surface per second
= 2.3×10
15
×0.5/100
= 1.15×10
13
electrons/s
The electron current = 1.15×10
13
×1.6×10
−19
A
= 1.84μA
Explanation:
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