Question

The momentum of an electron in an orbit is $\frac{h}{\lambda }$ where $h$ is plank's constant and $\lambda$ is wavelength associated with it. The nuclear magneton of electron of charge $e$ and mass $m_e$ is given as ${\mu }_n=\frac{eh}{3672\pi m_e}$. The dimensions of ${\mu }_n$ is

• A. $\left[M{L}^{2}A\right]$
• B. $\left[M{L}^{3}A\right]$
• C. $\left[L^{2}A\right]$
• D. $\left[M{L}^2\right]$

Answer 1

The correct option is C $\left[L^{2}A\right]$

$P=\frac{h}{\lambda }$
$\therefore {\mu }_n=\frac{eP\lambda }{3672\pi m_e}$
$\therefore \left[{\mu }_n\right]=\frac{\left[AT\right]\left[M{L}^{-1}\right]\left[L\right]}{\left[M\right]}=L^{2}A$