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(A) The energy of the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. When th

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(A) The energy of the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. When the atom is excited by $8.4 \ eV$,the new energy level $E_{new} = -13.6 + 8.4 = -5.2 \ eV$. Since $-5.2 \ eV$ does not correspond to any allowed energy level of the hydrogen atom (e.g.,$E_2 = -3.4 \ eV$,$E_3 = -1.51 \ eV$),the atom cannot be excited to a higher state by this specific energy. Therefore,no transition occurs,and the number of spectral lines emitted is $0$.

If a hydrogen atom is excited by an energy of $8.4 \ eV$,what will be the number of spectral lines emitted?

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