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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = \frac{E_1}{n^2}$,where $E_1$ is the ground state ene

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$n = 4$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = \frac{E_1}{n^2}$,where $E_1$ is the ground state energy. Given that the ionization energy $(I.P.)$ is $960 \ eV$,the ground state energy $E_1 = -960 \ eV$. Substituting the values into the formula: $-60 \ eV = \frac{-960 \ eV}{n^2}$. Solving for $n^2$: $n^2 = \frac{960}{60} = 16$. Therefore,$n = \sqrt{16} = 4$.

Suppose the ionization energy of a hydrogen-like species is $960 \ eV$. The value of the principal quantum number for an energy level of $-60 \ eV$ will be $......$.

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