Question

A stopping potential of 0.82V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength$4000\overset{\text{0}}{\mathop{\text{A}}}\,$. For light of wavelength$3000\overset{\text{0}}{\mathop{\text{A}}}\,$, the stopping potential is 1.85V. If the value of Planck’s constant is$6.5\times {{10}^{-X}}$. Find X?
[1 electron volt (eV)=$1.6\times {{10}^{19}}J$]

Answer 1

Hint : To solve this problem, we need to know the photoelectric effect, when light of frequency$\left( v \right)$ falls on a metal surface having work function (W) and stopping potential (V), then: $hv=W+eV$. Further, we will use the relation of frequency and wavelength, given by: $v=\dfrac{c}{\lambda }$.

Complete step by step solution :
To solve this problem, let’s first understand the photoelectric effect. The photoelectric is a phenomenon of ejection of electrons by a metal surface due to the surface being exposed to the electromagnetic spectrum. The ejected electrons are known as photoelectrons and the process of emission of electrons is known as photoemission.
For the photoelectric effect phenomenon to take place, light of a certain minimum frequency must be incident on the metal surface. The energy of the light wave is given by:$E=hv$, these energetic photons eject the loosely packed electrons at the metal’s surface.
These photoelectrons carry kinetic energy given by the excess energy given to the electron by the light wave. This kinetic energy of the photoelectron is given by the difference of the energy of the incident light and work function (W) of the metal. That is: $K.E=E-W\Rightarrow K.E=hv-W$.
The kinetic energy of the photoelectrons are found by applying positive potential to the plate towards which the photoelectrons are incident. This positive potential is known as stopping potential (V) as it literally does that, it stops those electrons giving us the information about the kinetic energy.
Therefore, the formula of the photoelectric effect becomes: $hv=W+eV\Rightarrow \dfrac{hc}{\lambda }=W+eV$.
Hence, $\dfrac{hc}{{{\lambda }_{1}}}=W+e{{V}_{1}}\to (1)$ and $\dfrac{hc}{{{\lambda }_{2}}}=W+e{{V}_{2}}\to (2)$. Subtracting equation(2) from (1), we get: $\dfrac{hc}{{{\lambda }_{2}}}-\dfrac{hc}{{{\lambda }_{1}}}=e{{V}_{2}}-e{{V}_{1}}\Rightarrow hc\left( \dfrac{1}{{{\lambda }_{2}}}-\dfrac{1}{{{\lambda }_{1}}} \right)=e\left( {{V}_{2}}-{{V}_{1}} \right)\to (3)$. The value of the work potential is constant, since the same metal is being exposed to different light waves of different wavelengths.
Here, from the problem we have:${{\lambda }_{1}}=4000\overset{0}{\mathop{\text{A}}}\,=4\times {{10}^{-7}}m$,${{V}_{1}}=0.82V$and${{\lambda }_{2}}=3000\overset{0}{\mathop{\text{A}}}\,=3\times {{10}^{-7}}m$,${{V}_{2}}=1.85V$. Further, we already know that, $c=3\times {{10}^{8}}\ m{{s}^{-1}}$.
Substituting in these values into equation (3), we get: $h\left( 3\times {{10}^{8}}\ \right)\left( \dfrac{1}{3\times {{10}^{-7}}}-\dfrac{1}{4\times {{10}^{-7}}} \right)=1.6\times {{10}^{-19}}\left( 1.85-0.82 \right)\Rightarrow h\left( 3\times {{10}^{8}}\ \right)\left( \dfrac{1}{12\times {{10}^{-7}}} \right)=1.6\times {{10}^{-19}}\left( 1.03 \right)$Therefore; $h=\dfrac{1.6\times {{10}^{-19}}\times 1.03\times 12\times {{10}^{-7}}}{3\times {{10}^{8}}\ }\Rightarrow h=6.592\times {{10}^{-34}}\ Js$.
Therefore, the value of X is 34.

Note : One must always remember that the photoemission only occurs when the light wave of at least threshold frequency or higher is incident on the metal surface. Hence, even if a highly intense lower frequency wave is incident on the metal, no photoemission of electrons will be observed in the photoelectric effect.