Modern Physics Ques 35
- As per Bohr model, the minimum energy (in $eV$ ) required to remove an electron from the ground state of doubly ionized Li atom $(Z=3)$ is
(1997, 1M)
(a) $1.51$
(b) $13.6$
(c) $40.8$
(d) $122.4$
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Answer:
Correct Answer: 35.(d)
Solution:
Formula:
- For hydrogen and hydrogen like atoms
$ E _n=-13.6 \frac{\left(Z^{2}\right)}{\left(n^{2}\right)}$ $ eV $
Therefore, ground state energy of doubly ionised lithium atom $(Z=3, n=1)$ will be
$ E _1=(-13.6) \frac{(3)^{2}}{(1)^{2}}=-122.4$ $ eV $
$\therefore$ Ionisation energy of an electron in ground state of doubly ionised lithium atom will be $122.4 $ $eV$.
BETA
