Question

The work function of cesium metal is 2.14 eV. When the light of frequency $6\times {{10}^{14}}Hz$ is incident on the metal surface, photoemission of electrons occurs. Calculate-
(a) The maximum kinetic energy of the emitted electron.
(b) Stopping potential
(c) The maximum speed of the emitted electrons.

Answer 1

Hint: To find the maximum kinetic energy, we can use the formula K = hv -$\varphi $. For the stopping potential, we can use the formula K = e${{V}_{0}}$and lastly for the maximum speed of the emitted electron, we can use the formula of kinetic energy, which is $K=\dfrac{1}{2}m{{v}^{2}}$.

Complete answer:
The work function is denoted as ${{\varphi }_{0}}$ and for cesium metal it is given as 2.14 eV.
The frequency of emission given to us is, v=$6\times {{10}^{14}}Hz$
Firstly, to find the kinetic energy of the emitted electron, we can use the formula,
K = hv - $\varphi $
Where, ‘K’ is the kinetic energy,
‘h’ is the Planck’s constant and the value of h is $h=6.625\times {{10}^{-34}}Js$
And v is the frequency of the emitted light, ${{\varphi }_{0}}$ is the work function.
Therefore, putting the values in the above formula, we get-$K=\left( \dfrac{6.625\times {{10}^{-34}}Js\times 6\times {{10}^{14}}Hz}{1.6\times {{10}^{-19}}C} \right)-2.14=0.344eV$, $1.6\times {{10}^{-19}}$C is the charge of one electron.
Therefore, the maximum kinetic energy of the emitted electron is 0.344 eV.
Next, we have to find out the stopping potential.
For stopping potential, we can use the formula which related kinetic energy to the stopping potential which is- K = e${{V}_{0}}$
Where, K is the kinetic energy, ‘e’ stands for the charge on one electron and ${{V}_{0}}$is the stopping potential.
Now we will put the values in the above equation to get the stopping potential-
$\begin{align}
  & K=e{{V}_{0}} \\
 & \Rightarrow {{V}_{0}}=\dfrac{K}{e}=\dfrac{0.345}{1.6\times {{10}^{-19}}}eV\text{ } \\
 & \text{or, }{{\text{V}}_{0}}=\dfrac{0.345}{1.6\times {{10}^{-19}}}\times 1.6\times {{10}^{-19}}V=0.345V \\
\end{align}$
Therefore, the stopping potential of the electron is 0.345 V.
And lastly we have to calculate the maximum speed of the emitted electron. As the electron is moving, we will use the kinetic energy formula-$K=\dfrac{1}{2}m{{v}^{2}}$
Where, K is the kinetic energy, ‘m’ is the mass of the electron and ‘v’ is the maximum speed.
Putting the values in the above equation, we will get-
     $\begin{align}
  & {{v}^{2}}=\dfrac{2K}{{{m}_{e}}} \\
 & \Rightarrow {{v}^{2}}=\dfrac{2\times 0.345\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}=1.213\times {{10}^{11}} \\
 & \Rightarrow v=\sqrt{1.213\times {{10}^{11}}}=3.483\times {{10}^{5}}m/s=348.3km/s \\
\end{align}$
Therefore, the maximum speed of the emitted electron is 348.3 km/s.

Note: The minimum frequency which is required for the emission of electrons from the metal surface is known as the threshold frequency. The stopping potential is the potential which is required to stop the ejection of electrons from the metal surface and it changes with the change in frequency. If the frequency of the beam is higher, the kinetic energy of the electron increases which will require a higher stopping potential.