Question

In an hydrogen atom, an electron is moving in the $n^{th}$ orbit. The circumference S of the orbit and the de-Broglie wavelength $\lambda$ of the moving electron are related by the equation:

• A. $S=n\lambda$
• B. $S=n/\lambda$
• C. $S=\lambda /n$
• D. $Sn\lambda =1$

Answer 1

Answer: (A)

$S=n\lambda$

The angular momentum formula :

$L=mvr---\left(1\right)$

And since for $H-$atom possible values of

$L=\frac{nh}{2\pi }---\left(2\right)$

$\therefore$ From $\left(1\right)$ and $\left(2\right)$

$\frac{nh}{2\pi }=mvr---\left(3\right)$

and according to de-broglie equation

$\lambda =\frac{h}{mv}$

$\Rightarrow mv=\frac{h}{\lambda }---\left(4\right)$

Putting $\left(4\right)$ in $\left(3\right)$

$\frac{nh}{2\pi }=\frac{hr}{\lambda }$

$2\pi r=n\lambda$

Hence; $n\lambda =2\pi r$

Where $2\pi r=$ Circumference $\left(S\right)$

$\Rightarrow S=n\lambda$