Question

In H-atom if $x$ is the radius of the first Bohr orbit, the De Broglie wavelength of an electron in 3rd orbit is:

• A. $3\pi x$
• B. $6\pi x$
• C. $\frac{9x}{2}$
• D. $\frac{x}{2}$

Answer 1

The correct option is B $6\pi x$

For a hydrogen atom, radius of nth orbit, $r_n=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z}$

Therefore,$\frac{r_1}{r_3}=\frac{1^2}{3^2}$

$r_3=9r_1$=$9x$

Now, according to De Broglie; angular momentum of electron in 3rd orbit is:
$mv{r}_3=\frac{3h}{2\pi }$ or $\frac{h}{mv}=\frac{2\pi r_3}{3}$
And
$\lambda =h/mv$ , where λ ,is the de Broglie wavelength.

Therefore,
$\lambda =\frac{2\pi r_3}{3}$
= $\frac{2\pi .9x}{3}$
=$6\pi x$

Hence the answer is 6$\pi x$.