Chapter 38 - Quantization - Exercises and Problems - Page 1115: 28

The radius of the atom is $~~1.90~nm$

We can find the state $n$: $E_n = \frac{13.6~eV}{n^2}$ $n^2 = \frac{13.6~eV}{E_n}$ $n = \sqrt{\frac{13.6~eV}{E_n}}$ $n = \sqrt{\frac{13.6~eV}{0.378~eV}}$ $n = 6$ We can find the radius: $r_n = n^2~a_B$ $r_6 = (6)^2~(0.0529\times 10^{-9}~m)$ $r_6 = 1.90\times 10^{-9}~m$ $r_6 = 1.90~nm$ The radius of the atom is $~~1.90~nm$