Modern Physics 1 Question 27
24. 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
Show Answer
Solution:
- 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 $122.4 eV$.
