The ionisation potential of hydrogen atom is | vedclass

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

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$3.4$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the second orbit,$n = 2$.Substituting the value of $n$ into the formula: $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.The energy required to remove an electron from the second orbit is the energy needed to bring the electron from $E_2 = -3.4 \ eV$ to $E = 0 \ eV$ (infinity).Therefore,the required energy is $0 - (-3.4) = 3.4 \ eV$.

The ionisation potential of hydrogen atom is $13.6 \ V$. The energy required to remove an electron from the second orbit of hydrogen is......$eV$.

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