Question

Photoelectric effect takes place in element $A$. Its work function is $2.5eV$ and threshold wavelength is $\lambda $. Another element $B$ is having a work function of $5eV$. Then find out the wavelength that can produce photoelectric effect in $B$.
Option A: $\dfrac{\lambda }{2}$
Option B: $2\lambda $
Option C: $\lambda $
Option D: $3\lambda $

Answer 1

Hint : Photoelectric effect is defined as the emission of electrons from the surface of metals after being hit by an electromagnetic radiation. The energy of these electromagnetic radiations are a property of their wavelengths.

Complete step by step answer:
When the surface of metal gets hit by an electromagnetic radiation, like light, of sufficient energy, the electrons attached to the surface of the metal get excited after absorbing that energy and get free. This effect is known as the photoelectric effect. The energy that the electromagnetic radiations carry, they carry in small packets. Those packets of energy are known as quanta. One quantum of light is known as a photon. The value of energy of a photon can be given by,
${E_P} = h\nu = \dfrac{{hc}}{\lambda }$
Where,
${E_P} = $Energy of the photon
$h = $Planck’s constant$ = 6.626 \times {10^{ - 34}}Js$
$\nu = $Frequency of the photon
$c = $Speed of light$ = 3 \times {10^8}m{s^{ - 1}}$
$\lambda = $Wavelength of the photon
The minimum energy required to just release the electron from the surface is known as the work function $\left( {{\phi _O}} \right)$ of the metal. When one photon of light hits the metal with energy equal to the work function, one electron absorbs that photon and leaves the surface, but with no energy because the energy provided by the photon is used up in freeing the electron. The wavelength of the light required for creating energy equal to the work function is known as threshold wavelength. When the photon has energy more than the work function, then the energy with which the electron leaves the surface is given by,
$E = \dfrac{{hc}}{\lambda } - {\phi _O}$
In this case, the energy provided by the photon is equal to the work function so the energy with which electron will leave is zero, so
$\dfrac{{hc}}{\lambda } - {\phi _O} = 0$
$ \Rightarrow {\phi _O} = \dfrac{{hc}}{\lambda }$
$ \Rightarrow {\phi _O} \propto \dfrac{1}{\lambda }$
$ \Rightarrow \dfrac{{{{\left( {{\phi _O}} \right)}_A}}}{{{{\left( {{\phi _O}} \right)}_B}}} = \dfrac{{{\lambda _B}}}{{{\lambda _A}}}$
$ \Rightarrow \dfrac{{2.5}}{5} = \dfrac{{{\lambda _B}}}{\lambda }$
$ \Rightarrow \dfrac{1}{2} = \dfrac{{{\lambda _B}}}{\lambda }$
$\therefore {\lambda _B} = \dfrac{\lambda }{2}$

Therefore the correct answer is option A.

Note: The metal with lowest work function is Cesium, and usually Cesium is used in most of the applications of photoelectric effect. Like photoelectric cells, photodiodes etc.