Question:
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 \AA$. The radius of the atom in the excited state, in terms of Bohr radius $\mathrm{a}_{0}$, will be: $(h c=12500 \mathrm{eV}-A)$
Correct Option: , 3
Solution:
(3) Energy of photon $=\frac{\text { hc }}{\lambda}=\frac{12500}{980}=12.75 \mathrm{eV}$
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit is given by
$E n=\frac{-13.6}{n^{2}} \Rightarrow E_{n}-E_{1}=-13.6\left[\frac{1}{n^{2}} \frac{-1}{1^{2}}\right]$
$\Rightarrow 12.75=13.6\left[\frac{1}{1^{2}} \frac{-1}{n^{2}}\right] \Rightarrow n=4$
$\therefore$ Electron will excite to $\mathrm{n}=4$
We know that ' $R$ ' $\propto n$ '
$\therefore$ Radius of atom will be $16 \mathrm{a}_{0}$