# Nuclear Magnetic Moments - Oxygen 15

• BOAS
In summary, the person is trying to calculate the nuclear magnetic moment for Oxygen 15 and their answer differs from the value in literature. They use equations to find the nuclear g-factor and ultimately find a value closer to the expected value after realizing a mistake in their calculations.
BOAS

## Homework Statement

I am trying to calculate the nuclear magnetic moment for Oxygen 15 but my answer differs strongly from the value quoted in literature.

## The Attempt at a Solution

[/B]
##\mu_J = g_J \mu_N J##

This isotope of Oxygen has an unpaired Neutron with ##J = \frac{1}{2}##.

Calculating the nuclear g-factor for ##J = l + \frac{1}{2}##

##g_J = g_l (1 - \frac{1}{2J}) + g_s(\frac{1}{2J})##

For a neutron ##g_l = 0## and so in this case ##g_J = g_s = -3.82##

I therefore find that ##\mu_J = -1.91 \mu_N## which is significantly different from the value I find in the NIST table of ##\pm 0.7189 \mu_N##

http://nist.gov/data/PDFfiles/jpcrd85.pdf

Is my treatment too simplistic to get a decent prediction, or am I making a fundamental error?

BOAS said:

## Homework Statement

I am trying to calculate the nuclear magnetic moment for Oxygen 15 but my answer differs strongly from the value quoted in literature.

## The Attempt at a Solution

[/B]
##\mu_J = g_J \mu_N J##

This isotope of Oxygen has an unpaired Neutron with ##J = \frac{1}{2}##.

Calculating the nuclear g-factor for ##J = l + \frac{1}{2}##

##g_J = g_l (1 - \frac{1}{2J}) + g_s(\frac{1}{2J})##

For a neutron ##g_l = 0## and so in this case ##g_J = g_s = -3.82##

I therefore find that ##\mu_J = -1.91 \mu_N## which is significantly different from the value I find in the NIST table of ##\pm 0.7189 \mu_N##

http://nist.gov/data/PDFfiles/jpcrd85.pdf

Is my treatment too simplistic to get a decent prediction, or am I making a fundamental error?

I think I have found my mistake.

My case is for J = L - S, with L = 1.

My g-factor is therefore ##g_J = g_l (1 + \frac{1}{2l + 1}) - g_s (\frac{1}{2l + 1})##, with ##g_l = 0##

I find that ##g_J = -0.63 \mu_N## which is much closer to my expected value.

Sorry for posting prematurely.

## 1. What is the significance of nuclear magnetic moments?

Nuclear magnetic moments are important because they provide information about the structure and behavior of atomic nuclei. They can also be used to study the properties of different elements and isotopes.

## 2. How is the nuclear magnetic moment of Oxygen 15 measured?

The nuclear magnetic moment of Oxygen 15 is measured using a technique called nuclear magnetic resonance (NMR). This involves placing the isotope in a strong magnetic field and measuring the energy levels of the nuclei as they absorb and emit electromagnetic radiation.

## 3. Why is Oxygen 15 a particularly interesting isotope to study?

Oxygen 15 is a radioactive isotope with a half-life of only 2.03 minutes, making it useful for medical imaging and other short-lived applications. It also has a unique nuclear spin and magnetic moment, providing valuable insights into the structure of atomic nuclei.

## 4. How does the nuclear magnetic moment of Oxygen 15 compare to other isotopes?

The nuclear magnetic moment of Oxygen 15 is relatively low compared to other isotopes. This is due to its high nuclear spin, which causes the magnetic moments of individual protons and neutrons to cancel each other out.

## 5. What are some practical applications of studying nuclear magnetic moments?

Studying nuclear magnetic moments can have a wide range of applications, from medical imaging and cancer treatment to materials science and environmental monitoring. It can also help us better understand the fundamental properties of matter and the forces that govern the behavior of atoms and molecules.

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