An electron in the hydrogen atom jumps excited state \[{\text{n}}\] to the ground state. The wavelength so emitted illuminates a photosensitive material having work function ${\text{2}}{\text{.75eV}}$. If the stopping potential of the photoelectron is ${\text{10V}}$, the value of ${\text{n}}$ is:

Hint:This question is based on the concept of photoelectric effect and the concept Bohr’s atom. It also involves the concept of stopping potential & Einstein’s Equations of photoelectric effect, along with the work function of the metal.

Complete step by step Solution:
From Einstein’s photoelectric effect equation.
Given as, Kinetic Energy maximum,
${\text{K}}{{\text{E}}_{{\text{max}}}}{{ = h\nu - w}}$
Where, ${{K}}{{{E}}_{{{max}}}}$ is the maximum kinetic energy of the emitted photoelectron
     ${{h = }}$ Plank’s constant
${{\nu = }}$ frequency of the incident photon
${{w = }}$ work function of the metal
It is given that work function,
${\text{w = 2}}{\text{.75ev}}$
It is known that the stopping potential.
${{\text{V}}_{{\text{stop}}}}{\text{ = 10V}}$
It is known that ${{\text{V}}_{{\text{stop}}}}$ is the potential which will stop even the fastest moving photoelectron. This implies that ${{\text{V}}_{{\text{stop}}}}{\text{ = K}}{{\text{E}}_{{\text{max}}}}$.
So, ${\text{e}}{{\text{V}}_{{\text{stop}}}}{\text{ = K}}{{\text{E}}_{{\text{max}}}}$
$ \Rightarrow {{\text{V}}_{{\text{stop}}}}{\text{ = }}\dfrac{{{\text{K}}{{\text{E}}_{{\text{max}}}}}}{{\text{e}}}$
If ${\text{K}}{{\text{E}}_{{\text{max}}}}$ is in Joules, then $\dfrac{{{\text{K}}{{\text{E}}_{{\text{max}}}}}}{{\text{e}}}$ should be in ${\text{eV}}$.
So , ${\text{K}}{{\text{E}}_{{\text{max}}}}\left( {{\text{eV}}} \right){\text{ = 10}}$
Putting the values in the photoelectric effect equation, ${{10 = h\nu - 2}}{\text{.75}}$
 ${{h\nu = 12}}{\text{.75eV}}$
Now, coming to the concept of Bohr atom, we know that ${{h\nu }}$ is the energy of the photon emitted when the electron jumps from ${{\text{n}}^{{\text{th}}}}$ orbit to the ground state.
Energy difference ${{\Delta }}{{\text{E}}_{\text{n}}}$ between ${{\text{n}}^{{\text{th}}}}$ excited state and ground state is given by,
${{\Delta E = 13}}{{.6 \times }}\left( {{{1 - }}\dfrac{{\text{1}}}{{{{\text{n}}^{\text{2}}}}}} \right)$
Energy difference ${{\Delta }}{{{E}}_{\text{n}}}{\text{ = 12}}{\text{.75eV}}$
So, putting the values, we get
${\text{12}}{\text{.75 = 13}}{{.6 \times }}\left( {{{1 - }}\dfrac{{{1}}}{{{{{n}}^{{2}}}}}} \right)$
$ \Rightarrow \dfrac{{{\text{12}}{\text{.75}}}}{{{\text{13}}{\text{.6}}}}{\text{ = 1 - }}\dfrac{{\text{1}}}{{{{\text{n}}^{\text{2}}}}}$
$ \Rightarrow {\text{0}}{\text{.9375 = 1 - }}\dfrac{{\text{1}}}{{{{\text{n}}^{\text{2}}}}}$
$ \Rightarrow \dfrac{{\text{1}}}{{{{\text{n}}^{\text{2}}}}}{\text{ = 0}}{\text{.0625}}$
$
   \Rightarrow {{\text{n}}^{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{0}}{\text{.0625}}}} \\
    \\
\Rightarrow {\text{n = }}\sqrt {\dfrac{{\text{1}}}{{{\text{0}}{\text{.0625}}}}} {\text{ = 4}}$

So, the electron drops from the state with, ${\text{n = 4}}$ to ground state.

Additional Information:Care should be taken in interpreting the units as most of the data is given in terms of ${\text{eV}}$. The conversion of ${\text{eV}}$into Joule or Vice versa should be remembered. Also it's a question having multiple concepts, so all the concepts should be remembered properly.

Note:If kinetic energy maximum has to be found out in ${\text{eV,}}$ then its value in ${\text{eV}}$ is equal is always equal to the value of stopping potential in Volts.