Question

If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be:
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)
(A)- $\dfrac{1}{2}\lambda$
(B)- $\dfrac{3}{4}\lambda$
(C)- $\dfrac{2}{3}\lambda$
(D)- $\dfrac{4}{9}\lambda$

Answer 1

The photoelectric effect takes place with the emission of the electron, when irradiated with the electromagnetic radiations. When the binding energy is negligible, then the intensity of the radiation falling on the metal surface leads to the ejection of electrons and the kinetic energy of the ejected electron is proportional to the photon energy.

Complete step by step answer:
The given phenomenon is known as the Photoelectric effect, where on striking the metal surface with the photons from the electromagnetic radiations, emission of electrons takes place.
The photon energy irradiated over the surface is partly used for breaking away the electrons from the surface and the remaining is used in the kinetic energy of the ejected electron. Thus, the relation for the kinetic energy of the ejected electron is given as:
$\,\text{Energy}\,\text{of}\,\text{EM}\,\text{radiation = Work function}+\text{Kinetic}\,\text{energy}$
$E=h{{\nu }_{0}}+K.E$
Then, given that the Kinetic energy $\gg$ Work function, so the above equation becomes:
$E=K.E$
$h\nu =\dfrac{1}{2}m{{v}^{2}}$ , where $\nu$ is the frequency of the radiation and $v$ is the velocity of the electron.
As, $\nu =c/\lambda$, the above equation becomes, $\dfrac{hc}{\lambda }=\dfrac{1}{2}m{{v}^{2}}$
And, as the momentum $P=mv$ , multiplying and dividing by m on the RHS of above equation, we get,
$\dfrac{hc}{\lambda }=\dfrac{1}{2}m{{v}^{2}}\times \left( \dfrac{m}{m} \right)=\dfrac{1}{2m}{{(mv)}^{2}}=\dfrac{{{P}^{2}}}{2m}$
Therefore, we get the relation between the momentum of the electron and the wavelength of the light as:
${{P}^{2}}\propto \dfrac{1}{\lambda }$
Given that, the wavelength of the light is${{\lambda }_{1}}=\lambda$, then the momentum is p. So, having the momentum as 1.5 p, the wavelength ${{\lambda }_{2}}$will be obtained as follows:
${{\left( \dfrac{{{P}_{1}}}{{{P}_{2}}} \right)}^{2}}=\dfrac{{{\lambda }_{2}}}{{{\lambda }_{1}}}$
${{\left( \dfrac{p}{1.5p} \right)}^{2}}=\dfrac{{{\lambda }_{2}}}{\lambda }$
${{\lambda }_{2}}={{\left( \dfrac{10}{15} \right)}^{2}}\lambda ={{\left( \dfrac{2}{3} \right)}^{2}}\lambda =\dfrac{4}{9}\lambda$

Therefore, for 1.5 p momentum of the photoelectron, the wavelength of the light will be option (D)- $\dfrac{4}{9}\lambda$.

Note: The work function is the amount of work done to remove an electron from the surface by breaking its bonds. Thus, also known as the binding energy of the electron. It is equal to the $h{{\nu }_{0}}$, where ${{\nu }_{0}}$ is the threshold frequency, that is the minimum frequency an electron should have in order to eject from the surface.