Question: Q. 9. In the first excited state of hydrogen atom, its radius is found to be $21.2 \times 10^{-11} \mathrm{~m}$. Calculate its Bohr radius in the ground state. Also calculate the total energy of the atom in the second excited state.
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Solution:
Ans.
$$ r_{0}=21.2 \times 10^{-11} \mathrm{~m} $$
First excited state means, $n=2$
$$ \begin{equation*} r_{0}=n^{2} r_{1} \tag{1} \end{equation*} $$
or
$$ r_{1}=\frac{r_{0}}{n^{2}} $$
$$ r_{1}=\frac{21.2 \times 10^{-11}}{4} \mathrm{~m} $$
$$ \begin{aligned} & =5.3 \times 10^{-11} \mathrm{~m} \ E & =-\frac{13.6}{n^{2}} \ E & =-\frac{13.6}{2^{2}}=-3.4 \mathrm{eV} \end{aligned} $$
