Question

A small piece of cesium metal $\left( {\phi = 1.9\,eV} \right)$ is kept at a distance of $20\,cm$ from a large metal plate having a charge density of $1.0 \times {10^{ - 9}}C\,{m^{ - 2}}$ on the surface facing the cesium piece. A monochromatic light of wavelength $400\,nm$ is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any chance in the electric field due to the small piece of cesium present.

Answer 1

Hint- We can find the potential of the metal plate as the product of electric field and distance. The potential of the metal surface can provide an energy equal to $qV$, where q is the charge and V is the potential of the plate. This will be the minimum energy required for an electron which is free at the cesium surface to reach the metal plate. The maximum kinetic energy will be the energy of the emitted electron with maximum kinetic energy from the cesium surfaces plus the energy provided by the potential on the metal surface on this electron.

Complete step by step solution:
It is given that a small piece of cesium metal is kept at a distance of $20\,cm$ from a large metal plate having a charge density $1.0 \times {10^{ - 9}}C\,{m^{ - 2}}$.
$d = 20\,cm$
$\Rightarrow d = 20\, \times {10^{ - 2}}m$
$\sigma = 1.0 \times {10^{ - 9}}C\,{m^{ - 2}}$
A monochromatic light of wavelength $400\,nm$ is incident on the cesium plate.
$\lambda = 400\,nm$
We need to find the minimum and maximum kinetic energy of the photoelectron reaching the large metal plate.
First, let us calculate the potential due to the large metal plate.
We know that potential is the product of electric fields and distance.
$V = E \times d$
Where, E is the electric field and d is the distance.
The electric field due to a large metal plate with surface charge density $\sigma$ is given as
$E = \dfrac{\sigma }{{{\varepsilon _0}}}$
Where, ${\varepsilon _0}$ is the permittivity of free space with value $8.85 \times {10^{ - 12}}\,{C^2}/N\,{m^2}$
Thus, the potential can be calculated as
$\Rightarrow V = \dfrac{\sigma }{{{\varepsilon _0}}} \times d$
$\Rightarrow V = \dfrac{{9 \times {{10}^{ - 9}}}}{{8.85 \times {{10}^{ - 12}}}} \times 20 \times {10^{ - 2}}$
$\Rightarrow V = 22.6\,\,V$
Now let us calculate the kinetic energy of the emitted electrons by using Einstein's photoelectric equation.
The photoelectric equation is given as
$h\upsilon = \phi + KE$
Where $h\upsilon$ denotes the energy of a photon, $\phi$ denotes the work function of a metal and KE denotes the kinetic energy.
$KE = h\upsilon - \phi$.................(2)
The energy of the photon can be calculated as
$E = h\upsilon$....................(1)
where, $\upsilon$ is the frequency, h is the Planck’s constant
$h = 6.626 \times {10^{ - 34}}\,J/s$
We know that, $c = \upsilon \lambda$ ,where c is the velocity of light, $\upsilon$ is the frequency and $\lambda$
$\Rightarrow \upsilon = \dfrac{c}{\lambda }$
On substituting the values in the equation 1 we get
$\Rightarrow E = h\dfrac{c}{\lambda }$
Thus equation 2 becomes,
$\Rightarrow KE = h\dfrac{c}{\lambda } - \phi$
$\Rightarrow KE = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{400 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} - 1.9eV$
$\Rightarrow KE = 1.205\,eV$
The energy due to potential is given as $qV$, where q is the charge and V is the potential. Here q is the charge of an electron. An electron which is made free at the surface of cesium metal will reach the metal plate with this minimum kinetic energy which is provided by the potential on the surface of the metal plate.
So, the minimum kinetic energy of the electron reaching the metal plate will be the energy provided by the potential difference that is
$\Rightarrow K{E_{\min }} = e \times V$
$\Rightarrow K{E_{\min }} = 22.6\,\,eV$
The maximum kinetic energy will be the sum of the maximum kinetic energy provided by the incident radiation to the emitted electron and the minimum energy provided by the potential of the plate.
$\Rightarrow KE = 22.6\,eV + 1.205\,eV$
$\Rightarrow K{E_{\max }} = 23.805\,eV$

Note: Remember that when an electron is free from the metal surface it can move towards the plate with the energy provided by the high potential of the metal plate. This will be the minimum energy with which it reaches the plate. To find the maximum kinetic energy we need to consider the minimum energy given by the plate and the maximum energy the emitted electron can have by the energy provided by the incident photon when it is accelerated in the direction of the metal plate.