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

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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 $E_n = -\frac{K}{n^2}$,where $K = 13.6 \ eV$.The energy difference between $n = 2$ and $n = 3$ orbits is $E = E_3 - E_2 = -\frac{K}{3^2} - (-\frac{K}{2^2}) = K(\frac{1}{4} - \frac{1}{9}) = K(\frac{5}{36})$.Therefore,$E = \frac{5}{36}K$,which implies $K = \frac{36}{5}E$.The energy required to remove an electron from the ground state $(n = 1)$ to infinity $(n = \infty)$ is the ionization energy: $E_{ion} = E_{\infty} - E_1 = 0 - (-\frac{K}{1^2}) = K$.Substituting the value of $K$ from the first equation: $E_{ion} = \frac{36}{5}E = 7.2 \ E$.

In a hydrogen atom,if the energy difference between the $n = 2$ and $n = 3$ orbits is $E$,how much energy (in terms of $E$) is required to remove an electron from the ground state?

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