# How do you calculate the magnetic moment of the ground state of Calciu

• zhillyz
In summary, to calculate the magnetic moment of the ground state of \,_{20}^{39}Ca, the nuclear shell model knowledge and the formula \mu_{n} = \left[g_{l}\dfrac{j(j+\dfrac{1}{2})}{(j+1)}-\dfrac{1}{2}\dfrac{1}{j+1}g_{s}\right]\mu_{N} can be used. The resulting magnetic moment was found to be μn=3.865x10^-27 J/T. However, for comparison to the experimental value in table 1, the magnetic moment should be converted to unitless units of magnetic moment/nuclear magneton, giving a value of 0
zhillyz

## Homework Statement

Calculate the magnetic moment of the ground state of $\,_{20}^{39}Ca$. Compare to the experimental value in table 1.

## Homework Equations

Nuclear Shell Model knowledge

## The Attempt at a Solution

Well firstly the magnetic moment of the nucleus similar to the spin is based only on the unpaired nucleon. With the above having one unpaired neutron in the 1d3/2 giving it spin 3/2 and parity +.

I thought the formulae for magnetic moment was:

$\mu_{n} = \dfrac{g_{n}\mu_{N}\vec{s}}{\hbar}$

but that would only involve googling the g value and multiplying it by 3/2 I think which would not get me all the marks this question is worth. Please give me a point in the right direction.

Last edited:
I thought the subshell it would be in would be 0d3/2

Yes sorry the subshell is 0d3/2 also its total angular momentum is 3/2 not its spin which I said before. I continued studying on my own when I didnt get an answer and in case anyone is interested the answer (I think) would be to use this formula (derived from the first equation I mentioned taking into account that s and l are not exactly defined in a system where j is, so replacing all terms in quantities related to j using vectors and expectation values);

$\mu_{n} = \left[g_{l}\dfrac{j(j+\dfrac{1}{2})}{(j+1)}-\dfrac{1}{2}\dfrac{1}{j+1}g_{s}\right]\mu_{N}$

for the magnetic moment i got μn=3.865x10^-27 J/T
where gl=0 and gs=-3.8261

That is what I got if you convert to those units. Though the comparison against the experimental value in table 1 is unitless i.e. magnetic moment/nuclear magneton and the equation above gives you an answer in units of nuclear magneton(if you don't put in the number of mu_N and just leave it as mu_N). So its just that number(0.76526)

## 1. How do you calculate the magnetic moment of the ground state of Calcium?

The magnetic moment of an atom is calculated by multiplying the number of unpaired electrons by the Bohr magneton value. For the ground state of Calcium, the number of unpaired electrons is 0, which gives a magnetic moment of 0.

## 2. What is the Bohr magneton value?

The Bohr magneton value is a physical constant equal to 9.274 x 10^-24 joule per tesla. It represents the magnetic moment of an electron in its ground state orbit around a nucleus.

## 3. How do you determine the number of unpaired electrons in the ground state of Calcium?

The number of unpaired electrons in the ground state of Calcium can be determined by looking at its electron configuration, which is 1s^22s^22p^63s^23p^64s^2. The last two electrons in the 4s orbital are paired, so there are no unpaired electrons.

## 4. Can the magnetic moment of the ground state of Calcium be affected by external factors?

Yes, the magnetic moment of an atom can be affected by external factors such as an applied magnetic field, temperature, and chemical bonding. These factors can cause the electrons to become excited and result in a non-zero magnetic moment.

## 5. How does the magnetic moment of Calcium's ground state compare to other elements?

The magnetic moment of Calcium's ground state is relatively low compared to other elements, as it has a full outer electron shell and no unpaired electrons. Other elements with unpaired electrons, such as Iron and Cobalt, have higher magnetic moments.

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