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Question: A beam of light of wavelength \(400nm\) and power of \(1.55mW\) is directed at the cathode of a phot...

A beam of light of wavelength 400nm400nm and power of 1.55mW1.55mW is directed at the cathode of a photoelectric cell. If only 10%10\% of the incident photons effectively produce a photoelectron, then find current due to these electrons. (Given hc=1240eVnmhc = 1240eVnm, e=1.6×10−19Ce = 1.6 \times {10^{ - 19}}C)
A)5μA5\mu A
B)40μA40\mu A
C)50μA50\mu A
D)114μA114\mu A

Explanation

Solution

Hint First we need to know the number of photons in the beam. Number of photons (NN) is a beam of power (PP) is given by
P=N×EP = N \times E
Where EE is the energy of the beam. Only 10%10\% of NN photons produce photoelectrons. One photon produces one photoelectron. So, the number of photoelectrons ejected is also 10%10\% of NN. Current produced by them will be the product of 10%10\% of NN and their charge ee.

Complete step-by-step solution :
We are given the wavelength of the incident beam , λ=400nm\lambda = 400nm and power as 1.55mW1.55mW.
We have the relation between power of incident beam and number of photons as
P=N×EP = N \times E
Where EE is the energy of the beam.
Energy of photons is given by
E=hcλE = \dfrac{{hc}}{\lambda }
Substituting the values, we have
1.55×10−3=N×1240×1.6×10−19400 N=1.55×10−3×4001240×1.6×10−19 =3.125×1015  1.55 \times {10^{ - 3}} = N \times \dfrac{{1240 \times 1.6 \times {{10}^{ - 19}}}}{{400}} \\\ N = \dfrac{{1.55 \times {{10}^{ - 3}} \times 400}}{{1240 \times 1.6 \times {{10}^{ - 19}}}} \\\ = 3.125 \times {10^{15}} \\\
N=3.125×1015N = 3.125 \times {10^{15}}photons per second.
Now, only 10%10\% of NN photons emit photoelectrons, so,
10%10\% of NN =10100××3.125×1015=3.125×1014 = \dfrac{{10}}{{100}} \times \times 3.125 \times {10^{15}} = 3.125 \times {10^{14}} photons per second.
Since, we know that one photon emits one photoelectron, so
3.125×10143.125 \times {10^{14}} photons will emit the same number of photoelectrons.
Current due to these electrons will be the product of the number of electrons and their charge ee.
Current is
I=3.125×1014×1.6×10−19 =5×10−5A =50×10−6A =50μA  I = 3.125 \times {10^{14}} \times 1.6 \times {10^{ - 19}} \\\ = 5 \times {10^{ - 5}}A \\\ = 50 \times {10^{ - 6}}A \\\ = 50\mu A \\\
So, option C is correct.

Note:-
You must be careful while doing calculations in large exponents. You must also be comfortable in unit conversions as in the given question, instead of using eVeV as a unit we have converted it in Joules by multiplying with 1.6×10−191.6 \times {10^{ - 19}}.