The electron in the hydrogen atom jumps from excited state $(n=3)$ to its ground state $(n=1)$ and the…
The electron in the hydrogen atom jumps from excited state $(n=3)$ to its ground state $(n=1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \mathrm{eV}$, the stopping potential is estimated to be (the energy of the electron in $n$th state $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^2} \mathrm{eV}$ )
$5.1 \mathrm{~V}$
$12.1 \mathrm{~V}$
$17.2 \mathrm{~V}$
$7 \mathrm{~V}$
Solution
For $n=1, \mathrm{E}_1=-\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}$ and for $n=3, E_3=-\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}$
So, required energy
$\begin{array}{l}
\mathrm{E}=\mathrm{E}_3-\mathrm{E}_1 =-(1.51)-(-13.6) =12.09 \mathrm{eV} \\
\because \mathrm{E} =\mathrm{W}+\mathrm{eV} \\
\therefore \mathrm{eV} =\mathrm{E}-\mathrm{W} \\
\mathrm{eV} =(12.09-5.1) \mathrm{e} =7 \text { volt }
\end{array}$