Question

According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is ${10}^{–9}\text{m}$. The principle quantum number for this electron is

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is B $3$

Given:
$\lambda ={10}^{-9}\text{m}$
Radius of hydrogen atom is given by
$r=0.529n^2\text{Å}$
In the orbit of hydrogen atom the de-Broglie wavelength is given by
$\lambda =2\pi rn$
$\Rightarrow n=\frac{\lambda }{2\pi r}=\frac{{10}^{-9}}{2\times 3.14\times 5.13\times {10}^{-11}}=3$

Hence, option $\left(C\right)$ is correct.