Answer
The radius of the atom is $~~0.476~nm$
Work Step by Step
We can find the de Broglie wavelength:
$\lambda = \frac{h}{m~v}$
$\lambda = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)(7.3\times 10^5~m/s)}$
$\lambda = 1.0\times 10^{-9}~m$
$\lambda = 1.0~nm$
We can find the state $n$:
$2\pi~r_n = n~\lambda$
$2\pi~n^2~a_B = n~\lambda$
$n = \frac{\lambda}{2\pi~a_B}$
$n = \frac{1.0\times 10^{-9}~m}{(2\pi)~(0.0529\times 10^{-9}~m)}$
$n = 3$
We can find the radius of the atom:
$r_n = n^2~a_B$
$r_3 = (3)^2~(0.0529\times 10^{-9}~m)$
$r_3 = 0.476\times 10^{-9}~m$
$r_3 = 0.476~nm$
The radius of the atom is $~~0.476~nm$
