Question

The de Broglie wavelength of an electron in the 3rd Bohr orbit is

• A. $2\pi a_0$
• B. $4\pi a_0$
• C. $6\pi a_0$
• D. $8\pi a_0$

Answer 1

Answer: (C)

We know
$2\pi r=n\lambda$
$n=3$
and $r=a_0\frac{n^2}{Z}$
$2\pi a_0\frac{n^2}{Z}=n\lambda$
$2\pi \frac{3^2{a}_0}{1}=3\lambda$
$\lambda =6\pi a_0$