The binding energy of a hydrogen atom in the | vedclass

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(B) The binding energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{13.6 \ eV}{n^2}$.For the ground s

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$13.6, 3.4, 1.5$

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(B) The binding energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{13.6 \ eV}{n^2}$.For the ground state $(n = 1)$:$E_1 = \frac{13.6}{1^2} = 13.6 \ eV$.For the first excited state $(n = 2)$:$E_2 = \frac{13.6}{2^2} = \frac{13.6}{4} = 3.4 \ eV$.For the second excited state $(n = 3)$:$E_3 = \frac{13.6}{3^2} = \frac{13.6}{9} \approx 1.51 \ eV$.Thus,the energies required to remove the electron from these levels are $13.6 \ eV, 3.4 \ eV,$ and $1.5 \ eV$ respectively.

The binding energy of a hydrogen atom in the ground state is $13.6 \ eV$. The energies required to remove an electron from the first three energy levels of the hydrogen atom are,respectively,(in $eV$):

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