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Q.
In hydrogen atom, if the difference in the energy of the electron in $n=2$ and $n=3$ orbits is $E$, the ionization energy of hydrogen atom is
Atoms
Solution:
Energy, $E=K\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right](K=$ constant $)$
$n_{1}=2 $ and $ n_{2}=3 $ So $ E=K\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=K\left[\frac{5}{36}\right]$
For removing an electron $n_{1}=1$ to $n_{2}=\infty$
Energy $E_{1}=K[1]=\frac{36}{5} E=7.2 E$
$\therefore$ Ionization energy $=7.2 E$