Question

The de-Broglie's wavelength of an electron in the $4^{\text{th}}$ orbit is ______ $\pi a_0$. ($a_0 = \text{Bohr's radius}$)

Answer 1

The condition for the de-Broglie wavelength of an electron in an orbit is given by:

$2 \pi r_n = n \lambda_d$

where $r_n$ is the radius of the $n$-th orbit, $n$ is the principal quantum number, and $\lambda_d$ is the de-Broglie wavelength.

The radius of the $n$-th orbit in the Bohr model is given by:

$r_n = 2 \pi a_0 \frac{n^2}{Z}$

Substituting this into the equation:

$2 \pi a_0 \frac{n^2}{Z} = n \lambda_d$

For an electron in the 4th orbit and for a hydrogen atom ($Z = 1$):

$2 \pi a_0 \frac{4^2}{1} = 4 \lambda_d$

Simplifying:

$\lambda_d = 8 \pi a_0$

Hence, the de-Broglie's wavelength of the electron in the 4th orbit is $8 \pi a_0$.