Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1113: 35

$0.332nm$

We can find the de Broglie wavelength of an electron in the ground state as follows: $\lambda_n=\frac{2\pi r_n}{n}$ We plug in the known values to obtain: $\lambda_1=\frac{(2\pi r_1)}{1}$ $\lambda_1=2\pi(5.29\times 10^{-11})$ $\lambda=0.332nm$