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**Hint:**The ionization energy of the shell is the negative of the energy of the shell that is $I.E = - {E_1}$ The energy for the nth main shell is given by the following relation

${E_n} = \dfrac{{{E_1}}}{{{n^2}}} \times {Z^2}$

where Z is the atomic number, n is the orbit in which the electron is present. We will use this formula in the given question and arrive at the correct answer.

**Complete step by step solution:**

We have been given the ionization energy of hydrogen-like species that is $960\;eV$ so by the following relation $I.E = - {E_1}$ . So accordingly using the formula that is

Energy for an nth main shell of hydrogen atom=Ionization energy of hydrogen atom /${n^2}$

So putting the given values in the equation, we get

$ - 60 = \dfrac{{ - 960}}{{{n^2}}}$

$ \Rightarrow $ $60 = \dfrac{{960}}{{{n^2}}}$

${n^2} = \dfrac{{960}}{{60}} = 16 \Rightarrow n = 4$

Therefore the value of principal quantum number having energy equal to $ - 60$ eV is four

**So, the correct answer is Option C.**

**Additional information:**

For hydrogen-like species, the energy of an electron in the nth orbit is given by ${E_n} = R\dfrac{{{z^2}}}{{{n^2}}}J/atom$ where R is the Rydberg constant which is equal to $2.18 \times {10^{ - 18}}$ .

Also, the ionization energy increases with an increase in atomic number and decreases with a decrease in atomic number due to a decrease in proximity and attractive force with the nucleus.

**Note:**ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. For hydrogen-like species, Z is taken to be one and the ionization energy of the species is the negative of the energy of the electron in the ground state.

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