Question:
The de Broglie wavelength of an electron in the $4^{\text {th }}$ Bohr orbit is:
Correct Option: , 4
Solution:
$2 \pi r=n \lambda$
$r=\frac{n^{2} a_{0}}{Z}$
$2 \pi \times \frac{4^{2}}{1} a_{0}=4 \lambda$
$\lambda=2 \pi \times \frac{4}{1} a_{0}$
$\lambda=8 \pi a_{0}$