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

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$7.2$

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.The energy difference between $n=2$ and $n=3$ is given as $E = E_3 - E_2$.$E = -\frac{13.6}{3^2} - \left( -\frac{13.6}{2^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight)$.Thus,$13.6 = E 	imes \frac{36}{5} = 7.2 E$.The ionization energy of a hydrogen atom is the energy required to move an electron from $n=1$ to $n=\infty$,which is $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 	ext{ eV}$.Therefore,the ionization energy is $7.2 E$.

In a hydrogen atom,if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E,$ the ionization energy of the hydrogen atom is ....... $E$.

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