Question

(a) Suppose you put both electrons in a helium atom into the $n=2$state;

what would the energy of the emitted electron be?

(b) Describe (quantitatively) the spectrum of the helium ion,$\mathit{H}{\mathit{e}}^{\mathbf{+}}$.

Short answer

(a)The energy of the emitted electron is $\mathbf{2}{\mathit{E}}_{\mathbf{1}}\mathbf{-}\mathbf{4}{\mathit{E}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathit{E}}_{\mathbf{1}}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{2}\mathit{e}\mathit{V}$

(b)Helium has one electron and it’s a hydrogenise ion with z=2 so the spectrum

is$\mathbf{1}\mathbf{/}\mathit{\lambda }\mathbf{=}\mathbf{4}\mathit{R}\left(1/n_f^2-1/n_i^2\right)$

Step-by-step solution

(a) The energy of the emitted electron

The energy of each electron is$E=Z^2{E}_1/n^2=4E_1/4=E_1=E_1=-13.6eV$,

so the total initial energy is$2\times \left(-13.6\right)eV=-27.2eV$.

One electron drops to the ground state $Z^2{E}_1/1=4E_1$, so the other is left

with $2E_1-4E_1=-2E_1=27.2eV$.

(b) The spectrum of the helium ion

(b) $H{e}^{+}$has one electron; it’s a hydrogenise ion with Z = 2, so the spectrum

is $1/\lambda =4R\left(1/n_f^2-1/n_i^2\right)$, where R is the hydrogen Rydberg constant, and $ni,nf$are the

initial and final quantum numbers (1, 2, 3, . . . ).