Question

The circumference of $n$th orbit in $H$-atom can be expressed in terms of de Broglie wavelength $\lambda$ as:

• A. $\left(0.529\right)n\lambda$
• B. $\sqrt{n\lambda }$
• C. $13.6\lambda$
• D. $n\lambda$

Answer 1

The correct option is D $n\lambda$

Accordinng to de-Broglie,
$\lambda =\frac{h}{mu}$
and circumference of nth orbit is
$mur=\frac{nh}{2\pi }$ (according to Bohr model)
$\therefore 2\pi r=nh/mu=n\lambda ,$
the circumference of an orbit is integer multiple of $\lambda$.