Question

1 BM is equal to:
A.$\dfrac{{he}}{{4mc\pi }}$
B.$\dfrac{{hc}}{{4\pi m}}$
C.$\dfrac{{{e^2}hc}}{{4m}}$
D.$\dfrac{{ehc}}{{\pi m}}$

Answer 1

Bohr Magneton is the unit of magnetic moment. It is a physical constant. Magnetron is the smallest value of magnetic moment. Magnetic moment is the magnetic strength and the orientation of a magnet that produces magnetic fields.

Complete step-by-step answer: BM is equal to $\dfrac{{eh'}}{{2{m_e}c}}$, where e is the charge, h’ is the reduced Planck constant, ${m_e}$ is the mass of the electron at rest and c is the speed of light.
h’ = $\dfrac{h}{{2\pi }}$
Hence 1 BM = $\dfrac{{eh}}{{4\pi {m_e}c}}$
The above formula is defined in Gaussian CGS units. The Bohr Magneton is the magnitude of magnetic dipole moment of an electron orbiting in an atom with such angular momentum.
The magnetic moment of an electron is either caused by its orbital or spin angular momentum. The value of Bohr Magneton in SI units is ${\rm{9}}{\rm{.27 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}{\rm{ J}}{{\rm{T}}^{{\rm{ - 1}}}}$.
Bohr Magneton is the smallest value of magnetic moment. Magnetic moment is the magnetic strength and the orientation of a magnet that produces magnetic fields.
Magnetic momentum of charged particles can be generated by two ways. The first way is that the moving electric charge forms a current, hence, due to this, the orbital motion is generated. This magnetic moment is generated by Ampere’s circuital law. The second way is that the electron possesses the spin magnetic moment.

Therefore, the correct option is A.

Note: In Bohr’s atomic model, a natural unit for the orbital angular momentum was denoted h’.
The Planck constant relates a photon’s energy to its frequency as well as it is the quantum of electromagnetic action. A photon’s energy is equal to the product of the Planck constant with the photon’s frequency. In quantum mechanics, the Planck constant is of fundamental importance.