Question #157797

Calculate ionization potential and first excitation potential of the hydrogen

atom taking ℎ = 6.62 x 10−34𝐽𝑠, 𝑒 = 1.6 x 10−19

, 𝑚 = 9.1 x 10−31𝑘𝑔.


1
Expert's answer
2021-01-24T14:23:26-0500

The ionization potential of the hydrogen atom is the energy required to remove one electron at the ground state from the atom. We can use Rydberg equation:


1λ=R(1n121n2)=Rn12, E0=hν=hcλ=hRc=2.181018 J, or 13.6 eV.\frac1\lambda=R\bigg(\frac{1}{n_1^2}-\frac{1}{n_\infty^2}\bigg)=\frac{R}{n_1^2},\\\space\\ E_0=h\nu=h\frac c\lambda=hRc=2.18\cdot10^{-18}\text{ J, or 13.6 eV.}

The first excitation potential of the hydrogen atom is


V=34E0=3413.6=10.2 eV.V=\frac{3}{4}E_0=\frac34\cdot13.6=10.2\text{ eV}.




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