Question

The de Broglie wavelength of an electron in $2^{nd}$ orbit is ( $a_0$ = Bohr radius )

• A. $2\pi a_{0}$
• B. $4\pi a_{0}$
• C. $8\pi a_{0}$
• D. $16\pi a_{0}$

Answer 1

Answer: (B)

Radius of $n^{th}$ orbit is
$r_n = n^{2} a_0$ where $a_0$ is, the Bohr radius.
If $\lambda$ is the de Broglie wavelength of an electron while revolving in nth orbit of radius $r_n$, then
$2\pi r_{n} = n\lambda; \quad\lambda = \frac{2\pi r_{n}}{n}$
For $2^{nd}$ orbit, $n = 2$
$\therefore \quad \lambda = \frac{2\pi r_{n}}{n} = \frac{2\pi\left(2\right)^{2} a_{0}}{2} = 4\pi a_{0}$