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

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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 $E_n = -\frac{E_o}{n^2}$,where $E_o = 13.6 \ eV$.The energy difference between $n = 2$ and $n = 3$ is given as $E = E_3 - E_2 = -\frac{E_o}{3^2} - (-\frac{E_o}{2^2}) = E_o(\frac{1}{4} - \frac{1}{9}) = \frac{5E_o}{36}$.Therefore,$E_o = \frac{36E}{5} = 7.2E$.The ionization potential of the $H$ atom is the energy required to remove an electron from $n = 1$ to $n = \infty$,which is $E' = E_{\infty} - E_1 = 0 - (-\frac{E_o}{1^2}) = E_o$.Substituting the value of $E_o$,we get $E' = 7.2E$.

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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