Question

In a photoemissive cell with exciting wavelength $\lambda $, the fastest electron has speed v. If the exciting wavelength is changed by $\dfrac{3\lambda }{4}$, the speed of the fatest emitted electron will be:
$\text{A}\text{. }v{{\left( \dfrac{3}{4} \right)}^{\dfrac{1}{2}}}$
$\text{B}\text{. }v{{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}$
$\text{C}\text{. less than }v{{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}$
$\text{D}\text{. greater than }v{{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}$

Answer 1

Hint: Use the theory of photoelectric effect. Use the equation for the maximum kinetic energy of a photo electron i.e. $\dfrac{1}{2}mv_{\max }^{2}=h\dfrac{c}{\lambda }-\phi $. Write the expressions for the maximum speeds when the wavelengths are $\lambda $ and $\dfrac{3\lambda }{4}$ respectively. Divide both the velocities to find the relation between them.

Formula used:
$E=h\nu $
$\nu =\dfrac{c}{\lambda }$
${{K}_{\max }}=E-\phi $
${{K}_{\max }}=\dfrac{1}{2}mv_{\max }^{2}$

Complete step by step answer:
According to Einstein’s photoelectric theory, when a light is incident on a piece of metal, electrons are emitted from the surface of the metal.
It is considered that light consists of particles called photons. Each photon has an energy equal to $E=h\nu $ ….. (i) ,

where E is the energy of the photon, h is the Planck’s constant and $\nu $ is the frequency of the light.
The frequency of light is related with the wavelength of the light as $\nu =\dfrac{c}{\lambda }$, c is the speed of light. Substitute the value of frequency in equation (i).
$E=h\dfrac{c}{\lambda }$ …. (ii)

When a photon of light collides with an electron at the surface of the metal, the electron is ejected from the surface. The ejected electron consists of some kinetic energy. The maximum kinetic energy that the electron can have is given as ${{K}_{\max }}=E-\phi $ …. (iii).

$\phi $ is called the work function of the metal. It is the minimum energy required to eject the electron.
The kinetic energy of the electron will be ${{K}_{\max }}=\dfrac{1}{2}mv_{\max }^{2}$
Substitute the value of E and ${{K}_{\max }}$ in equation (iii).

$\dfrac{1}{2}mv_{\max }^{2}=h\dfrac{c}{\lambda }-\phi $.
It is given that when the wavelength is $\lambda $ the maximum speed of an electron is v.
Therefore,

$\dfrac{1}{2}m{{v}^{2}}=h\dfrac{c}{\lambda }-\phi $ ….. (iv).
When the wavelength of light is $\dfrac{3\lambda }{4}$, let the maximum speed of an electron be v’.
Therefore,

$\dfrac{1}{2}mv{{'}^{2}}=h\dfrac{c}{\dfrac{3\lambda }{4}}-\phi $ ….. (v).
Divide equation (v) by equation (iv).
 $\Rightarrow \dfrac{\dfrac{1}{2}mv{{'}^{2}}}{\dfrac{1}{2}m{{v}^{2}}}=\dfrac{h\dfrac{c}{\dfrac{3\lambda }{4}}-\phi }{h\dfrac{c}{\lambda }-\phi }$
$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{h\dfrac{4c}{3\lambda }-\phi }{h\dfrac{c}{\lambda }-\phi }$

Now add and subtract $\dfrac{4\phi }{3}$ in the numerator on the right side of the equation.

$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{h\dfrac{4c}{3\lambda }-\phi +\dfrac{3\phi }{4}-\dfrac{3\phi }{4}}{h\dfrac{c}{\lambda }-\phi }$
$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{\left( \dfrac{4hc}{3\lambda }-\dfrac{4\phi }{3} \right)+\left( -\phi +\dfrac{4\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi }$
$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{\dfrac{4}{3}\left( \dfrac{hc}{\lambda }-\phi \right)+\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi }$
$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{\dfrac{4}{3}\left( \dfrac{hc}{\lambda }-\phi \right)}{\dfrac{hc}{\lambda }-\phi }+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi }$
$\Rightarrow \dfrac{v{{'}^{2}}}{{{v}^{2}}}=\dfrac{4}{3}+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi }$
$\Rightarrow v{{'}^{2}}=\left( \dfrac{4}{3}+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi } \right){{v}^{2}}$
$\Rightarrow v'={{\left( \dfrac{4}{3}+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi } \right)}^{\dfrac{1}{2}}}v$
Here, $\dfrac{hc}{\lambda }-\phi $ is a positive value according to equation (iv) and $\dfrac{\phi }{3}$ is also positive.
Hence, $\left( \dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi } \right)$ is a positive value.
Therefore, $\dfrac{4}{3}+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi }$ > $\dfrac{4}{3}$.
This implies that ${{\left( \dfrac{4}{3}+\dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi } \right)}^{\dfrac{1}{2}}}$ > ${{\left( \dfrac{\left( \dfrac{\phi }{3} \right)}{\dfrac{hc}{\lambda }-\phi } \right)}^{\dfrac{1}{2}}}$

This further implies that v’ > ${{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}v$.
Therefore, the fastest electron emitted when the wavelength of light is $\dfrac{3\lambda }{4}$ is greater than ${{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}v$.
Hence, the correct option is D.

Note:
If a photon having energy greater than the work function collides with an electron, then the electron will eject from the surface of the metal. In the process, the other electrons influence the photoelectron. Hence, the photoelectron will have an energy less than the maximum kinetic energy.