Question

The de Broglie wavelength of an electron in the 4th Bohr orbit is:

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

Answer 1

The correct option is D $8\pi a_0$

de Broglie wavelength $\left(\lambda \right)$
$2\pi r=n\lambda$
$n\text{is integer}$
$r\text{raduis of orbit}$
$Z\text{is atomic number}$
$n=4\text{\&}r=a_0\frac{n^2}{Z}2\pi a_0\frac{n^2}{z}=n\lambda$
$2\pi \frac{4^2{a}_1}{1}=4\lambda$
$\lambda =8\pi a_0$