Question

Find the radius and energy of a $H{e}^{+}$ ion in the states

(a) n = 1,

b) n = 4 and

(c) n = 10

Answer 1

(a) Here n = 1, r = $\frac{{ϵ}_0{h}^2{n}^2}{\pi mZ{e}^2}$

= $\frac{0.53n^2}{z}{A}^0$

$ϵ=\frac{-13.6Z^2}{n^2}=\frac{13.6\times 4}{4}$

= -54.4 eV

(b) n = 4, r = $\frac{0.53\times 16}{2}=4.24A^0$

$ϵ=\frac{-13.6\times 4}{164}=-3.4eV$

(c) n = 10, r = $\frac{0.53\times 100}{2}$

= $26.5A^0$

$ϵ=\frac{-13.6\times 4}{100}=-0.544eV$