Question

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

Answer 1

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

 Complete step-by-step solution:
We are given the wavelength of the incident beam , $\lambda = 400nm$ and power as $1.55mW$.
We have the relation between power of incident beam and number of photons as
$P = N \times E$
Where $E$ is the energy of the beam.
Energy of photons is given by
$E = \dfrac{{hc}}{\lambda }$
Substituting the values, we have
$
  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}} \\
 $
$N = 3.125 \times {10^{15}}$photons per second.
Now, only $10\% $ of $N$ photons emit photoelectrons, so,
$10\% $ of $N$$ = \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
$3.125 \times {10^{14}}$ 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 $e$.
Current is
$
  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 $eV$ as a unit we have converted it in Joules by multiplying with $1.6 \times {10^{ - 19}}$.