Understanding Magnetic Moments: Bohr and Nuclear

In atomic and nuclear physics, magnetic moments represent the magnetic strength and orientation of a particle or atom. Two fundamental units used to express these magnetic moments are the Bohr magneton and the nuclear magneton.

Defining the Bohr Magneton ($\mu_B$)

The Bohr magneton is the smallest possible magnetic moment that an electron can possess due to its intrinsic angular momentum (spin). It is defined as:

$$ \mu_B = \frac{e\hbar}{2m_e} $$

Where:

• e is the elementary charge (the magnitude of the charge of an electron).
• \hbar (h-bar) is the reduced Planck constant ($\frac{h}{2\pi}$).
• m_e is the rest mass of the electron.

The Bohr magneton is primarily used to describe the magnetic moments associated with electrons orbiting the nucleus.

Defining the Nuclear Magneton ($\mu_N$)

Similarly, the nuclear magneton is a unit used to measure the magnetic moments of particles found within the nucleus, such as protons and neutrons. It is defined using the mass of the proton:

$$ \mu_N = \frac{e\hbar}{2m_p} $$

Where:

• e is the elementary charge.
• \hbar is the reduced Planck constant.
• m_p is the rest mass of the proton.

Because the proton is much heavier than the electron (m_p > m_e), the nuclear magneton is much smaller than the Bohr magneton.

Calculating the Ratio

The question asks for the ratio of the nuclear magneton to the Bohr magneton. Let's calculate this ratio:

$$ \text{Ratio} = \frac{\mu_N}{\mu_B} $$

Substitute the definitions:

$$ \text{Ratio} = \frac{\frac{e\hbar}{2m_p}}{\frac{e\hbar}{2m_e}} $$

We can simplify this expression by cancelling out the common terms ($e$, $\hbar$, and 2):

$$ \text{Ratio} = \frac{1/m_p}{1/m_e} = \frac{m_e}{m_p} $$

The Value of the Ratio

The ratio of the nuclear magneton to the Bohr magneton is equal to the ratio of the electron's mass to the proton's mass.

Experimentally, the mass of a proton is approximately 1836 times the mass of an electron:

$$ m_p \approx 1836 \times m_e $$

Therefore, the ratio is:

$$ \frac{\mu_N}{\mu_B} = \frac{m_e}{m_p} \approx \frac{m_e}{1836 \times m_e} = \frac{1}{1836} $$

Thus, the nuclear magneton is approximately 1/1836 times the Bohr magneton.