Question

Light of wavelength 2200 $\overset{\circ }{\mathop{A}}\,$ (angstrom) falls on a photosensitive plate with work function 4.1 eV. Find (a) energy of a photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.

Answer 1

Hint: Energy of a photon is given as $E=h\nu $. Use the relation between speed of light, its frequency and its wavelength to find the energy of a photon in eV. Maximum kinetic energy of the photoelectron is given as ${{K}_{\max }}=hv-\phi $.

Formula used:
$E=h\nu $
${{K}_{\max }}=hv-\phi $
$\nu =\dfrac{c}{\lambda }$
$1eV=1.6\times {{10}^{-19}}J$
W = qV

Complete step-by-step answer:
It is considered that light is made up of small energy packets called photons. You can think of a photon as a particle with no mass. However, a photon has momentum. The energy of a photon is given as $E=h\nu $, where h is Planck’s constant and $\nu $ is frequency of light.
When a beam of light falls on a metal, the photons collide with the electrons at the surface of the metal and transfer their energy to the electrons. If the electrons absorb more than a certain amount of energy (called work function), they get free from the electrostatic force and come out of the metal with some kinetic energy. This electron is called a photoelectron.
The maximum kinetic energy of an electron is given as ${{K}_{\max }}=hv-\phi $,
where $hv$ is the energy of the photon and $\phi $ is the amount of energy needed to break the bond between the electron and the nucleus.
Let us now calculate the things that are asked in the question.

(a) Energy of a photon is $E=h\nu $.
 Let us assume that this experiment is taking place in vacuum then,
$\nu =\dfrac{c}{\lambda }$, where c is the speed of light in vacuum and $\lambda $ is the wavelength of light.
Therefore,
$E=h\dfrac{c}{\lambda }$.
Here, $h=6.62\times {{10}^{-34}}Js$, $c=3\times {{10}^{8}}m{{s}^{-1}}$ and $\lambda =2200\overset{\circ }{\mathop{A}}\,=2200\times {{10}^{-8}}m$
Therefore,
$E=6.62\times {{10}^{-34}}Js.\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{2200\times {{10}^{-8}}m}=0.9\times {{10}^{-18}}J$ ……(i).s
$1eV=1.6\times {{10}^{-19}}J\Rightarrow 1J=\dfrac{1}{1.6\times {{10}^{-19}}}eV$
Substitute the value of 1J in equation (i).
$E=0.9\times {{10}^{-18}}.\dfrac{1}{1.6\times {{10}^{-19}}}eV=5.6eV$
Hence, the energy of the photons is 5.6eV.

(b) Maximum kinetic energy of the photoelectron is
 ${{K}_{\max }}=hv-\phi $
We just found the value of hv, which is equal to 5.6eV and it is given that $\phi =4.1eV$.
Therefore,
${{K}_{\max }}=5.6-4.1=1.5eV$.
Hence, the maximum kinetic energy of the electron is 1.5eV.

(c) Stopping potential is a potential difference that is created to stop the motion electron.
Here the electron has a kinetic energy of 1.5eV. Therefore, a negative work must be done on the electron to bring it to rest.
Let that potential difference be P. Since, W=qV, if the electron travels through a potential difference of P, a work will be done on it equal to eP.
Therefore, ${{K}_{\max }}=eP=1.5eV$
$\Rightarrow P=1.5V$.
Hence, the stopping potential in this case is 1.5V.

Note: One photon can eject only one electron at a time. And the electron will eject out of the metal only if the energy of that photon is more than the work function of the metal.
Not every photoelectron possesses a kinetic energy equal to ${{K}_{\max }}$, because there are a lot of factors that influence the motion of the electron, such as repulsion due to other electrons, electron collide with each other dissipating some amount of energy.