Question

The de-Broglie wavelength of electron in second Bohr's orbit is equal to ____________.

• A. circumference of the orbit
• B. half the circumference of the orbit
• C. twice the circumference of the orbit
• D. four times circumference of the orbit

Answer 1

The correct option is B half the circumference of the orbit

Conservation of angular momentum gives:
$mvr=\frac{nh}{2\pi }\Rightarrow \frac{h}{mv}=\frac{2\pi r}{n}$

De Broglie's wavelength, $\lambda =\frac{h}{mv}$

Thus, for $n=2$

$\Rightarrow \lambda =\frac{2\pi r}{2}=\frac{circumference}{2}$