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  4. The ionization energy of the hydrogen atom is 13.6 eV. The potential energy of the electron in n = 2 state of hydrogen atom is

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Q. The ionization energy of the hydrogen atom is $13.6\, eV$. The potential energy of the electron in $n = 2$ state of hydrogen atom is

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Solution:

Given, $E_{n}=13.6\, e V$
Energy of an electron in $n^{\text {th }}$ state
$E_{n}=\frac{-13.6 z^{2} e V}{n^{2}}$
$\therefore $ Energy of an electron in $n=2$ state So,
$E_{2}= \frac{18-6 z^{2}}{(2)^{2}}=-3.4 \,e V $
$P E =2 E_{n=2}$
$=2 \times(-3.4)$
$ \approx-6.8 \,eV$