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

Vedclass · Accepted Answer

$7.2$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \, eV$. Given that the energy difference between $n = 3$ and $n = 2$ is $E$: $E = E_3 - E_2 = \left( \frac{-13.6}{3^2} ight) - \left( \frac{-13.6}{2^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight) \, eV$. So,$E = \frac{13.6 	imes 5}{36} \, eV$. The ionization potential of the $H$ atom is the energy required to remove an electron from $n = 1$ to $n = \infty$,which is $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \, eV$. From the expression for $E$,we have $13.6 = \frac{36E}{5} = 7.2E$. Therefore,the ionization potential is $7.2E$.

One energy difference between the states $n = 2$ and $n = 3$ is $E \, eV$ in a hydrogen atom. The ionization potential of the $H$ atom is ........... $E$.

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