The electron in a hydrogen atom,upon absorbin | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n_1 = 1)$,the

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$4$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \ eV$.When the electron absorbs energy $\Delta E = 12.75 \ eV$,it jumps to a higher orbit $n_2$.The energy of the final state is $E_{n_2} = E_1 + \Delta E = -13.6 + 12.75 = -0.85 \ eV$.Using the formula $E_{n_2} = -\frac{13.6}{n_2^2}$,we get:$-0.85 = -\frac{13.6}{n_2^2}$$n_2^2 = \frac{13.6}{0.85} = 16$$n_2 = 4$.Therefore,the electron jumps to the $4^{th}$ orbit.

The electron in a hydrogen atom,upon absorbing $12.75 \ eV$ of energy,will jump to the ....... orbit.

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