Question

For hydrogen atom electron in $n$ th Bohr orbit, the ratio of radius of orbit to its de-Brogile wavelength is :

• A. $\frac{n}{2\pi }$
• B. $\frac{n^2}{2\pi }$
• C. $\frac{1}{2\pi n}$
• D. $\frac{1}{2\pi n^2}$

Answer 1

The correct option is A $\frac{n}{2\pi }$

In $n^{th}$ orbit of hydrogen atom,

$m{v}_n{r}_n=\frac{nh}{2\pi }$

Velocity, $v_n=\frac{nh}{2\pi m{r}_n}$

The de-Broglie wavelength, $\lambda =\frac{h}{m{v}_n}$

$\lambda =\frac{h}{m\left(\frac{nh}{2\pi m{r}_n}\right)}$

$\lambda =\frac{2\pi r_n}{n}$

$\frac{r_n}{\lambda }=\frac{n}{2\pi }$

The correct option is A.