Question

The de Broglie wavelength of an electron moving in a circular orbit is $\lambda .$ The minimum radius of orbit is:

• A. $\frac{\lambda }{\pi }$
• B. $\frac{\lambda }{2\pi }$
• C. $\frac{\lambda }{4\pi }$
• D. $\frac{\lambda }{3\pi }$

Answer 1

The correct option is B $\frac{\lambda }{2\pi }$

From de-Broglie equaiton,
$\lambda =\frac{h}{mv}\Rightarrow mv=\frac{h}{\lambda }$
For an allowed orbit, the angular momentum is $mvr=\frac{nh}{2\pi }$
$\Rightarrow \frac{h}{\lambda }.r=\frac{nh}{2\pi }$ $(\because mv=\frac{h}{\lambda })$
$\Rightarrow r=\frac{n\lambda }{2\pi }$
For minimum radius, n = 1
$\therefore r=\frac{\lambda }{2\pi }.$