Answer:
Option C
Explanation:
$ E_{n}=-2.18 \times 10^{-18}. \frac {Z^{2}}{n^{2}}$
where, En = energy of an electron in nth orbital.
Z= atomic number
n= number of orbital
For hydrogen , Z=1
Therefore,
$E_{3}=-2.18\times 10^{-18}\times\frac{1}{3^{2}}$
$=-\frac{2.18\times 10^{-18}}{9}$
=$-0.242 \times 10^{-18}$
and $E_{\infty}=-2.18 \times 10^{-18}\times \frac{1}{\infty}$
$E_{\infty}=0$
Hence, $E_{\infty}$=0 and $E_{3}=-0.242 \times 10^{-18}$
and option (c) is the correct answer.