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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ eV / n^2$. The energy difference between $n = 3$ and

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$7.2$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ eV / n^2$. The energy difference between $n = 3$ and $n = 2$ is $\Delta E = E_3 - E_2 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} ight) = -13.6 \left( \frac{1}{9} - \frac{1}{4} ight) = -13.6 \left( \frac{4 - 9}{36} ight) = -13.6 \left( -\frac{5}{36} ight) = \frac{13.6 	imes 5}{36} \ eV$. Given $\Delta E = E$,so $E = \frac{68}{36} \ eV = \frac{17}{9} \ eV$. The ionization energy of the $H$-atom is the energy required to move an electron from $n = 1$ to $n = \infty$,which is $E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$. Substituting $13.6 = E 	imes \frac{36}{5}$,we get $13.6 = E 	imes 7.2$. Therefore,the ionization energy is $7.2 \ E$.

In a hydrogen atom,the energy difference between the states $n = 2$ and $n = 3$ is $E \ eV$. The ionization energy of the $H$-atom is: (in $E$)

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