Question

One milliwatt of light of wavelength ${{\lambda = 4560}}{{{A}}^{{o}}}$ is incident on a cesium metal surface. Calculate the electron current liberated. Assume a Quantum efficiency of ${{n = 0}}{{.5\% }}$ [Work function of Cesium ${{ = 1}}{{.89eV}}$] take ${{hc = 12400eV - A}}$.

Answer 1

Hint: We calculate the number of electrons with the help of formula, and multiply it by electronic charge then we get a photoelectric current because current is basically a flow of electrons.
Formula Used:
${{E = }}\dfrac{{{{hc}}}}{{{\lambda }}}$ where ${{h}}$ is planck's constant
${{c}}$ is speed of light
${{\lambda }}$ is Wavelength
${\text{Photoelectric current}} = {\text{no. of electrons} \times \text{electronic charge}}$

Complete step by step answer:
We know that when light of a certain frequency is incident on the surface of metal the emit electrons. When these electrons flow they give rise to an electric current. This current is known as photoelectric current.
We know that
Energy of incident photon of wavelength ${{4560}}{{{A}}^{{o}}}$ on Cesium surface is given by
${{E = }}\dfrac{{{{hc}}}}{{{\lambda }}}{{ = }}\dfrac{{{{6}}{{.62 \times 1}}{{{0}}^{{{ - 34}}}}{{ \times 3 \times 1}}{{{0}}^{{8}}}}}{{{{4560 \times 1}}{{{0}}^{{{ - 10}}}}}}$
Since, ${{h = 6}}{{.62 \times 1}}{{{0}}^{{{ - 34}}}}{{Joule/sec }}$
${{c = 3 \times 1}}{{{0}}^{{8}}}{{{m}}}/{{{s}}}$
So,
    \[{{E = }}\dfrac{{{{6}}{{.62 \times 1}}{{{0}}^{{{ - 34}}}}{{ \times 3 \times 1}}{{{0}}^{{8}}}}}{{{{4560 \times 1}}{{{0}}^{{{ - 10}}}}}}{{ = 4}}{{.32 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}\]
Now, Energy of incident in terms of Joule, i. e.
${{1mW = 1}}{{{0}}^{{{ - 3}}}}{{{J}}}/{{{S}}}$
So, the No. of electrons in ${{1mV}}$of light is $\dfrac{{{{1}}{{{0}}^{{{ - 3}}}}}}{{{{4}}{{.32 \times 1}}{{{0}}^{{{ - 19}}}}}}{{ = 2}}{{.32 \times 1}}{{{0}}^{{{15}}}}{{{{Photon}}}}/{{{{second}}}}$
Now, we have given the Quantum Efficiency is ${{0}}{{.5\% }}$ i.e. only $0.5\% $ of incident photons releases photoelectrons.
Thus, Number of photoelectrons emitted from Cesium surface per second is equal to
$ \Rightarrow {{0}}{{.5 \times 2}}{{.32 \times 1}}{{{0}}^{{{15}}}}$
$ \Rightarrow \dfrac{{{{05}}}}{{{{100}}}}{{ \times 2}}{{.32 \times 1}}{{{0}}^{{{15}}}}$ which is equal to
$ \Rightarrow {{1}}{{.16 \times 1}}{{{0}}^{{{13}}}}$
Therefore, ${\text{Photoelectric current}} = {\text{no. of electrons} \times \text{electronic charge}}$
${{ = 1}}{{.16 \times 1}}{{{0}}^{{{13}}}}{{ \times 1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}{{A}}$
Charge on 1e is $1.6 \times 10^{ - 19}C$
${{I = 1}}{{.856 \mu A}}$

Hence, photoelectric current will be $I_{equal}$ to $1.856 \mu A$.

Note: Photocurrent is the electric current through a photosensitive device, such as photodiode, as the result of exposure to radiant power. The photocurrent may occur as a result of the photoelectric, photo emissive, or photovoltaic effect. The Photocurrent may be enhanced by internal gain caused by interaction among ions and photons under the influence of applied fields, such as colour in an available photodiode(AP ${{\Delta }}$).
The Highest (maximum) value of retarding photocurrent is called Saturation current. When the potential at which photo-current becomes Zero is called Cut-off Voltage.
Photocurrent is the electric current through a photosensitive device, & is calculated by finding the number of electrons flowing per second in a diode.