If the ionisation energy for the hydrogen ato | vedclass

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

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$10.2$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$, the energy is $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.The next higher state is the first excited state $(n=2)$.The energy of the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 	ext{ eV}$.The energy required to excite the atom from the ground state to the first excited state is $\Delta E = E_2 - E_1$.$\Delta E = -3.4 	ext{ eV} - (-13.6 	ext{ eV}) = -3.4 	ext{ eV} + 13.6 	ext{ eV} = 10.2 	ext{ eV}$.

If the ionisation energy for the hydrogen atom is $13.6$ eV, then the energy required to excite it from the ground state to the next higher state is nearly (in $eV$)

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