Question

The maximum wavelength of radiation that can produce photo-electric effect in a certain metal is $200nm$. The maximum kinetic energy acquired by electron due to radiation of wavelength $100nm$ will be:
$\begin{aligned} & \text{A}\text{. }12.4eV \\\ & \text{B}\text{. }6.2eV \\\ & \text{C}\text{. }100eV \\\ & \text{D}\text{. }200eV \\\ \end{aligned}$

Answer 1

When light falls on a metal plate, electrons are ejected from the metal surface with some energy. For finding the maximum kinetic energy of photoelectrons, we will use Einstein’s equation for the photoelectric emission. Einstein’s photoelectric emission explains the relation between the energy associated with emitted electrons and the wavelength of incident light.

Formula used:
Kinetic energy of emitted electrons,
$KE=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{o}}}$

Complete step by step answer:
The photoelectric effect is described as the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons being emitted in this manner are called photoelectrons. The photoelectric effect is often defined as the ejection of electrons from a metal plate when light falls on it.

The Einstein’s equation for photoelectric effect is given by,
$E=hv-W$
Where,
$E$ is the energy associated with the photoelectrons
$h$ is the Planck's constant
$v$ is the frequency of incident light
$W$ is the work function of the metal
The equation can be rewritten as,
$\begin{aligned} & E=hv-h{{v}_{o}} \\\ & E=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{o}}} \\\ \end{aligned}$
Where,
$v$ is the frequency of incident light
${{v}_{o}}$ is the threshold frequency
$\lambda$ is the wavelength of incident light
${{\lambda }_{o}}$ is the wavelength associated to threshold frequency, that is, threshold wavelength
Threshold frequency is the amount of minimum frequency below which the photoelectric emission is not feasible irrespective of the intensity of the incident radiation.
From photoelectric equation, we have,
$KE=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{o}}}$
Where,
$h$ is the Planck’s constant
$\lambda$ is the wavelength of incident light
${{\lambda }_{o}}$ is the wavelength associated to threshold frequency, that is, threshold wavelength
For converting the energy from Joules $\left( J \right)$ to electron volts $\left( eV \right)$,
$KE\left( eV \right)=\dfrac{hc}{\lambda e}-\dfrac{hc}{{{\lambda }_{o}}e}$
Where,
$e$ is the charge on an electron, that is,
We have,
$\begin{aligned} & h=6.67\times {{10}^{-34}}J \\\ & c=3\times {{10}^{8}}m{{s}^{-1}} \\\ & e=1.6\times {{10}^{-19}}C \\\ \end{aligned}$
Therefore,
$\begin{aligned} & \dfrac{hc}{e}=\dfrac{6.67\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}} \\\ & \dfrac{hc}{e}=1240eVnm \\\ \end{aligned}$
Now,
$KE\left( eV \right)=\dfrac{hc}{\lambda e}-\dfrac{hc}{{{\lambda }_{o}}e}$
Putting values,
$\begin{aligned} & \lambda =100nm \\\ & {{\lambda }_{o}}=200nm \\\ \end{aligned}$
We get,

& KE\left( eV \right)=1240\left( \dfrac{1}{100}-\dfrac{1}{200} \right) \\\ & KE\left( eV \right)=\dfrac{1240}{200} \\\ & KE\left( eV \right)=6.2eV \\\ \end{aligned}$$ The maximum kinetic energy acquired by an electron due to radiation of wavelength $100nm$ will be $6.2eV$. **Hence, the correct option is B.** **Note:** The amount of energy associated with the emitted photons is directly proportional to the photon’s electromagnetic frequency, or the frequency of incident light, and inversely proportional to the wavelength of incident light. The higher the frequency of the wave, the higher the energy associated with the photon. The longer the wavelength, the lower is the energy associated with the photon.