Question: Q. 6. (i) The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \mathrm{~m}$. Calculate its radius in $n=3$ orbit.
(ii) The total energy of an electron in the first excited state of the hydrogen atom is $-3.4 \mathrm{eV}$. Find out its (a) kinetic energy and (b) potential energy in this state.
A [Delhi Comptt. I, II, III 2014]
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Solution:
Ans. (i) Radius of orbit
$$ r_{n}=n^{2} r_{0} $$
where, $r_{0}$ is Bohr’s radius $=5.3 \times 10^{-11} \mathrm{~m}$ radius of $n=3$ orbit
$$ \begin{aligned} r_{3} & =(3)^{2} \times 5.3 \times 10^{-11} \mathrm{~m} \ & =47.7 \times 10^{-11} \mathrm{~m} \ & =4.77 \times 10^{-10} \mathrm{~m} \end{aligned} $$
(a) Kinetic energy, $K=\frac{e^{2}}{8 \pi \varepsilon_{0} r}=-$ Total energy
Hence Kinetic energy, $K=-(-3.4) \mathrm{eV}=3.4 \mathrm{eV} \quad \mathbf{1}$
(b) Potential energy, $P=-\frac{e^{2}}{4 \pi \varepsilon_{0} r}=2 \times$ total energy
$$ \begin{equation*} =-6.8 \mathrm{eV} \tag{1} \end{equation*} $$
