Question

In Bohrs orbit of hydrogen atom ${}^{\prime }{m}^{\prime }$ kg is mass of an electron and ${}^{\prime }{e}^{\prime }$ coulomb is the charge on it. The ratio (in SI units) of magnetic dipole moment to that of the angular momentum of electron is

• A. $e/2m$
• B. $e/m$
• C. $2e/m$
• D. $2e/3m$

Answer 1

The correct option is A $e/2m$

Angular momentum of electron $=$$mvr$
Magnetic dipole movement $=\frac{evr}{2}$
So, ratio $=\frac{e}{2m}$