Question

If the charge to mass ratio of an electron is 'A' C/kg, then the gyromagnetic ratio of an orbital electron in C/kg is

• A. $\frac{A}{4}$
• B. $A$
• C. $2A$
• D. $\frac{A}{2}$

Hint:

The gyromagnetic ratio of an orbital electron is fundamentally always exactly half of its specific charge value ($\frac{e}{2m}$). Remembering this relationship helps you avoid deriving the orbital magnetic dipole moment mechanics during an exam!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem provides the specific charge (charge-to-mass ratio, $\frac{e}{m_e}$) of an electron as a constant '$A$'. We need to express the gyromagnetic ratio of an atomically bound orbiting electron in terms of this constant.

Step 2: Detailed Explanation:
Let's look at the definitions of the quantities involved:

• The specific charge of an electron is defined as the ratio of its charge magnitude $e$ to its rest mass $m_e$: $$ \text{Specific Charge} = \frac{e}{m_e} = A $$

• The gyromagnetic ratio ($\gamma$) of an electron is defined as the ratio of its magnetic dipole moment ($\mu_l$) to its orbital angular momentum ($L$): $$ \gamma = \frac{\mu_l}{L} $$ From Bohr's atomic model derivations, for a circular electron orbit of radius $r$ with speed $v$, the current is $I = \frac{ev}{2\pi r}$ and the magnetic area is $\pi r^2$, giving $\mu_l = I \cdot \text{Area} = \frac{evr}{2}$. The angular momentum is $L = m_e v r$. Taking the ratio reveals a constant relationship: $$ \gamma = \frac{\frac{evr}{2}}{m_e vr} = \frac{e}{2m_e} $$
Substituting our given parameter $\frac{e}{m_e} = A$ into the gyromagnetic ratio formula: $$ \gamma = \frac{1}{2} \left(\frac{e}{m_e}\right) = \frac{A}{2} $$

Step 3: Final Answer:
The gyromagnetic ratio is $\frac{A}{2}$, which corresponds to option (D).