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# A beam of light of wavelength 400nm and power 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 $hc = 1240eV - nm$,$e = 1.6 \times {10^{ - 19}}C$(A) $5\mu A$(B) $40\mu A$(C) $50\mu A$(D) $11.4\mu A$

Last updated date: 13th Jun 2024
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Hint: The energy of an incident photon is given by the expression, $E = \dfrac{{hc}}{\lambda }$. This value gives the number of electrons photons which can be produced per second, which also gives the number of electrons emitted per second. This can be further used to calculate the value of photocurrent by the relation $I = e \times n$.
Formula used:
$E = \dfrac{{hc}}{\lambda }$
$\dfrac{c}{\lambda } = \upsilon$
$P = nE$
$I = e \times n$
Where, E is the energy of an incident photon
h is the Planck’s constant which is equal to $6.62 \times {10^{ - 34}}{m^2}kg/s$
c is the speed of light which is equal to $3 \times {10^8}m/s$
$\lambda$ is the wavelength of electromagnetic wave used
$\upsilon$ is the frequency
$e$ is the charge on electron which is equal to $1.6 \times {10^{ - 19}}C$
$I$ is the value of photocurrent.
P is the power
n is the number of photons produced per second

During the photoelectric effect, the photon strikes the atom with an energy E which is a function of its frequency.
$E = h\upsilon$
Since in the question, frequency is not given, the term $\dfrac{c}{\lambda } = \upsilon$ is used.
So the energy of the photon,
$E = \dfrac{{hc}}{\lambda }$
The value of hc is given $1240eV$
We know that $1eV = 1.6 \times {10^{ - 19}}J$
Therefore,
$hc = 1240 \times 1.6 \times {10^{ - 19}}J - nm$
($nm$refers to nanometers)
$hc = 1984 \times {10^{ - 19}}J - nm$
$E = \dfrac{{hc}}{\lambda } = \dfrac{{1984 \times {{10}^{ - 19}}J - nm}}{{400nm}} = 4.96 \times {10^{ - 19}}J$
Therefore the number of the photons is given by-
$n = \dfrac{P}{E}$
Where P is the power given in the question as, $1.55mW$
Writing in SI units-
$n = \dfrac{{1.55 \times {{10}^{ - 3}}}}{{4.96 \times {{10}^{ - 19}}}} = 3.125 \times {10^{15}}$
Number of photons used to produce the photoelectric current are only $10\%$so,
$n = 3.125 \times {10^{15}} \times \dfrac{{10}}{{100}}{s^{ - 1}}$
$n = 3.125 \times {10^{14}}$
This also the number of electrons which produce current,
Now the photocurrent ‘I’ can be calculated as-
$I = e \times n$
Given, $e = 1.6 \times {10^{ - 19}}C$
$I = 1.6 \times 3.125 \times {10^{14}} \times {10^{ - 19}}$
$I = 5 \times {10^{ - 5}}A$
\$I = 50 \times {10^{ - 6}}A

Therefore option (c) is correct.

Note: The energy of a photon can be written as a function of its wavelength but it is not the ideal expression because the wavelength of the light can be changed. When the light passes through a medium, its speed decreases significantly, as the frequency of the light remains constant, its wavelength changes. Similar effect occurs in the Doppler effect as well, where red-shift (increase in wavelength) and blue-shift (decrease in wavelength) is observed.