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Supposing the ionization energy of hydrogen-like species is $960$ eV. Find out the value of principal quantum number having energy equal to $ - 60$ eV
(A) $n = 2$
(B) $n = 3$
(C) $n = 4$
(D) $n = 5$

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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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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