Question: Q. 7. In the ground state of hydrogen atom, its Bohr radius is given as $5.3 \times 10^{-11} \mathrm{~m}$. The atom is excited such that the radius becomes $21.2 \times 10^{-11} \mathrm{~m}$. Find (i) the value of the principal quantum number and (ii) the total energy of the atom in this excited state.
A [OD South 2016]
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Solution:
Ans. (i)
$$ r=r_{0} n^{2} $$
$$ 21.2 \times 10^{-11}=5.3 \times 10^{-11} n^{2} $$
$$ \Rightarrow \quad n=2 $$
(ii)
$$ \begin{aligned} E & =\frac{-13.6 \mathrm{eV}}{n^{2}} \ & =\frac{-13.6 \mathrm{eV}}{2^{2}}=-3.4 \mathrm{eV} \end{aligned} $$
[Award $1 / 2$ mark if the student just writes $\left.E=E_{1} / 4\right]$
[CBSE Marking Scheme 2016]
