Question

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $\nu$. If the wavelength is changed to $\dfrac{3\lambda }{4}$, the speed of the fastest electron will be
$\begin{aligned} & A)>\nu {{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}} \\\ & B)<\nu {{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}} \\\ & C)=\nu {{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}} \\\ & D)=\nu {{\left( \dfrac{3}{4} \right)}^{\dfrac{1}{2}}} \\\ \end{aligned}$

Answer 1

Kinetic energy of emitted electron is related to the wavelength of incident radiation as well as the work function of the photocell. Kinetic energy of an emitted electron is also related to the mass and velocity of the emitted electron. Both these relations are combined together to determine the relation between velocities of emitted electrons in the first case and the second case.
Formula used:
$1)E=\dfrac{hc}{\lambda }-\phi$
$2)E=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
Kinetic energy of an emitted electron is related to the wavelength of incident radiation as well as the work function of photocell. This is mathematically expressed as:
$E=\dfrac{hc}{\lambda }-\phi$
where
$E$ is the kinetic energy of emitted electron
$\lambda$ is the wavelength of incident radiation
$h$ is the Planck’s constant
$c$ is the speed of light
$\phi$ is the work function of photocell
Let this be equation 1.
We also know that the kinetic energy of an emitted electron is related to its mass as well as velocity. This is mathematically expressed as:
$E=\dfrac{1}{2}m{{v}^{2}}$
where
$E$ is the kinetic energy of emitted electron
$m$ is the mass of emitted electron
$v$ is the velocity of emitted electron
Let this be equation 2.
Combining equation 1 and equation 2, we have
$E=\dfrac{hc}{\lambda }-\phi =\dfrac{1}{2}m{{v}^{2}}$
Let this be equation 3.
Coming to our question, we are given that a radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $\nu$. If the wavelength is changed to $\dfrac{3\lambda }{4}$, we are required to determine the speed of the fastest electron.
If ${{E}_{1}}$ denotes the kinetic energy of the fastest electron in the first case, then, using equation 3, ${{E}_{1}}$ is given by
${{E}_{1}}=\dfrac{hc}{{{\lambda }_{1}}}-\phi =\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}\Rightarrow \dfrac{hc}{\lambda }-\phi =\dfrac{1}{2}{{m}_{1}}{{\nu }^{2}}$
where
${{\lambda }_{1}}=\lambda$ is the wavelength of incident radiation in the first case (as provided)
$m$ is the mass of emitted electron
${{v}_{1}}=\nu$ is the velocity of emitted electron in the first case (as provided)
$h$ is the Planck’s constant
$c$ is the speed of light
$\phi$ is the work function of photocell
Let this be equation 4.
Similarly, if ${{E}_{2}}$ denotes the kinetic energy of the fastest electron in the second case, then, using equation 3, ${{E}_{2}}$ is given by
${{E}_{2}}=\dfrac{hc}{{{\lambda }_{2}}}-\phi =\dfrac{1}{2}m{{v}_{2}}^{2}\Rightarrow \dfrac{hc}{\left( \dfrac{3\lambda }{4} \right)}-\phi =\dfrac{1}{2}m{{v}_{2}}^{2}\Rightarrow \dfrac{4hc}{3\lambda }-\phi =\dfrac{1}{2}m{{v}_{2}}^{2}$
where
${{\lambda }_{2}}=\dfrac{3\lambda }{4}$ is the wavelength of incident radiation in the second case
$m$ is the mass of emitted electron
${{v}_{2}}$ is the velocity of emitted electron in the second case
$h$ is the Planck’s constant
$c$ is the speed of light
$\phi$ is the work function of photocell
Let this be equation 5.
Now, from equation 4 and equation 5, it is clear that
${{E}_{2}}>\dfrac{4}{3}{{E}_{1}}\Rightarrow \dfrac{1}{2}m{{v}_{2}}^{2}>\left( \dfrac{4}{3} \right)\dfrac{1}{2}m{{\nu }^{2}}\Rightarrow \dfrac{1}{2}m{{v}_{2}}^{2}>\dfrac{2}{3}m{{\nu }^{2}}\Rightarrow {{v}_{2}}^{2}>\dfrac{4}{3}{{\nu }^{2}}\Rightarrow {{v}_{2}}>{{\left( \dfrac{4}{3} \right)}^{\dfrac{1}{2}}}\nu$
Let this be relation 6.

So, the correct answer is “Option A”.

Note:
Photoelectric effect refers to the phenomenon by which electrons get emitted from the surface of a solid, when radiation such as light is incident on the surface of the solid. The solid surface can also be called a photocell. Work function of a photocell is nothing but the minimum thermodynamic energy required to remove an electron from the surface of a photocell, to the surroundings and away from its surface.