In an inelastic collision, an electron excite | vedclass

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n=1)$, $E_1 = -

Vedclass · Accepted Answer

$+1.51 \ eV$

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n=1)$, $E_1 = -13.6 \ eV$.For the $M$-shell $(n=3)$, the energy is $E_3 = -\frac{13.6 \ eV}{3^2} = -\frac{13.6}{9} \ eV \approx -1.51 \ eV$.To ionize the hydrogen atom from the $M$-state, the atom must be provided with enough energy to reach the ionization limit $(E_{\infty} = 0 \ eV)$.The energy required is $\Delta E = E_{\infty} - E_3 = 0 - (-1.51 \ eV) = +1.51 \ eV$.Therefore, the second electron must transfer at least $+1.51 \ eV$ of energy to the atom.

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