Atomic Structure 1 Question 38
14. The energy of an electron in the first Bohr orbit of $\mathrm{H}$-atom is $-13.6 \mathrm{eV}$. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are)
(1988)
(a) $-3.4 \mathrm{eV}$
(b) $-4.2 \mathrm{eV}$
(c) $-6.8 \mathrm{eV}$
(d) $+6.8 \mathrm{eV}$
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Answer:
Correct Answer: 14. (a)
Solution:
- Energy of electron in H-atom is determined by the expression:
$$ E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV} \quad \text { where, } \quad n=1,2,3, \ldots $$
In excited states, $E_{2}=-\frac{13.6}{4}=-3.4 \mathrm{eV}$
$$ E_{3}=-\frac{13.6}{9}=-1.51 \mathrm{eV} \text { etc. } $$
