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Question

Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio h/e is:

A
6.6×1015JsC1
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B
4.12×1015JsC1
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C
6.6×1034JsC1
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D
4.12×1034JsC1
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Solution

The correct option is A 4.12×1015JsC1
Given:-
λ1=546nm=546×109m
(Kmax)=0.18eV=(0.18×e)J
λ2=185nm=185×109m
Stopping potential,V=4.6V(Nigative potential)
Solution:-
Let us assume the work function of the photo cathode is ϕo.
In the first case,
(Kmax)=hcλ1ϕo
0.18e=hc549×109ϕo(1)
In the second\quad case,
(Kmax)2=hcλ2ϕo(2)
Here,
(Kmax)2=e×Stoppingpotential)
[kineticenergyofelectronareconvertedtotheelectricpotentialenergy.]
(Kmax)2=e×4.6
usinginequation(2)4.6e=hc185×109ϕo(3)
Subtracting equation (1) from equation (3)
(4.6e)(0.18e)=hc[1185×1091546×109]
4.42e=hc×0.00357×109
he=4.423×108×0.00357×109
he=4.127×1015JsC1

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