Video: Calculating the Number of Microstates in the State of an Atom

What is the total number of microstates in a ⁡𝐷 term of the ground state iron atom?

05:05

Video Transcript

What is the total number of microstates in a five 𝐷 term of the ground state iron atom?

Okay, so in this question, we’re looking at a five 𝐷 term and we’re trying to work out the total number of possible microstates in this five 𝐷 term. So what does a five 𝐷 term actually mean?

Well, we can recall that term symbols are written using the following conventions. The superscript, which in this case is five, represents two 𝑆 plus one, where 𝑆 is the total spin quantum number. And the large value, which in this case is 𝐷, represents 𝐿, where 𝐿 is the total orbital angular momentum quantum number.

However, with 𝐿, we don’t actually write the quantum number itself, but instead we write the letter that represents the orbital that we’re studying. So our value of 𝐿 could actually be zero or one or two or three or so on, but in the term symbol, we don’t write zero, one, two, or three. We actually write 𝑆 or 𝑃 or 𝐷 or 𝐹 or so on and so forth, where 𝐿 is equal to zero corresponds to 𝑆, 𝐿 is equal to one corresponds to 𝑃; you get the idea.

For the two 𝑆 plus one part, however, it’s a lot simpler. We actually use the numerical value of 𝑆. So now that we know that that’s how term symbols are written, we can calculate the values of 𝑆 and 𝐿 for this term. So in our case, we’ve got a five 𝐷 term, and this is equal to two 𝑆 plus one in the superscript and 𝐿 as the main character, which means that two 𝑆 plus one is equal to five and the numerical value of 𝐿 is the one which corresponds with 𝐷. So in other words, 𝐿 is equal to two, and we can write that down here.

We can also rearrange this first equation here to find out what 𝑆 is. When we do that, we find that 𝑆 is also equal to two. So for our five 𝐷 term, we find that 𝑆 is equal to two and 𝐿 is equal to two. In other words, the total spin quantum number is equal to two and the total orbital angular momentum quantum number is also equal to two. So why is this relevant?

Well, it’s because we’re trying to work out the number of microstates. And all the possible microstates are just all possible combinations of π‘š sub 𝑆 and π‘š sub 𝐿 for this term. Now π‘š sub 𝑆 is the spin projection quantum number, and π‘š sub 𝐿 is known as the magnetic quantum number. And we work out the possible values of π‘š sub 𝑆 and π‘š sub 𝐿 using the values of 𝑆 and 𝐿, because we can recall that π‘š sub 𝑆 can have a possible set of values starting from negative 𝑆 and going up in integer steps, so negative 𝑆 plus one, negative 𝑆 plus two, and so on and so forth, until we get to 𝑆 minus one and then finally positive 𝑆. And these are all the possible values of π‘š sub 𝑆.

And the same is true for π‘š sub 𝐿, except this time you start with negative 𝐿 and then go up in integer steps, negative 𝐿 plus one, so on and so forth, until we get to 𝐿 minus one and then finally positive 𝐿. And those are all the possible values of π‘š sub 𝐿.

So we know how to calculate all of the possible values of π‘š sub 𝑆 and π‘š sub 𝐿, and these depend on the values of 𝑆 and 𝐿. So in a five 𝐷 state where 𝑆 is equal to two and 𝐿 is equal to two, the possible values of π‘š sub 𝑆 are starting with negative 𝑆, which is negative two, and then go up in integer steps, so negative one and then zero and then one and then two, and we stop there because 𝑆 is equal to two and we stop at positive 𝑆. Similarly, with π‘š sub 𝐿, all of its possible values are starting with negative 𝐿, which is negative two, and then going up in integer steps and then stopping at positive 𝐿. So those are all the possible values of π‘š sub 𝐿.

Now the values of π‘š sub 𝑆 and π‘š sub 𝐿 are independent of each other, or in other words an electron can have any combination of the values of π‘š sub 𝑆 and π‘š sub 𝐿. So the total possible number of microstates is just all possible combinations of π‘š sub 𝑆 and π‘š sub 𝐿 added together. In other words, all possible microstates include π‘š sub 𝑆 and π‘š sub 𝐿 is equal to, let’s start with negative two, negative two. Then there’s negative two, negative one, and so on and so forth, and we can keep counting all of the possible values of π‘š sub 𝑆 combined with all possible values of π‘š sub 𝐿.

And an intuitive way to think about this is that we could start with one value of π‘š sub 𝑆 β€” let’s say negative two β€” and then switch between all values of π‘š sub 𝐿 β€” so that’s five different values. So that’s already five different possible microstates. Then we switch to the next value of π‘š sub 𝑆, which is negative one, and once again we go through all five possible values for π‘š sub 𝐿.

So now we’ve got another five possible microstates. So, so far, we have five states from when π‘šπ‘† is equal to negative two and another five from when π‘šπ‘† is equal to negative one. And we’ll get another five from when π‘šπ‘† is equal to zero and another five from when it’s one and another five from when it’s two. In other words, the total number of microstates possible is five times five, which turns out to be 25. And that is the total possible number of microstates in a five 𝐷 term. And so we’ve arrived at our final answer. The total number of microstates is 25.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.