Question

The de Broglie wavelength of an electron in the $4^{th}$ Bohr orbit is :

• A. $6\pi a_{0}$
• B. $2\pi a_{0}$
• C. $8\pi a_{0}$
• D. $4\pi a_{0}$

Answer 1

Answer: (C)

$2\pi r=n \lambda$
$for\, n=1, r=a_{0}$
$n=4, r=16a_{0}$
$So,\, 2\pi \times16a_{0}=4\times\lambda$
$\lambda=8\pi a_{0}$