The ionization potential of a hydrogen atom i | vedclass

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

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$3.4$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$, $E_1 = -13.6 	ext{ eV}$.The first excited state corresponds to $n=2$.The energy of the electron in the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 	ext{ eV}$.To ionize the atom from this state, we need to provide enough energy to bring the electron to an energy level of $0 	ext{ eV}$ (the continuum).Therefore, the required energy is $\Delta E = 0 - (-3.4 	ext{ eV}) = 3.4 	ext{ eV}$.

The ionization potential of a hydrogen atom is $13.6 \text{ eV}$. How much energy needs to be supplied to ionize a hydrogen atom in the first excited state (in $\text{ eV}$)?

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