Question

Question #2ab5b

Answer 1

Here's how you can do that.

Explanation:

Let's use a hydrogen atom as an example here.

You know that the atom is excited to the fourth excited level, which corresponds to `n=5` because `n=1` is the ground state.

Your goal here is to figure out how many transitions are possible between the ground state and the fourth excited level.

Notice that when energy is absorbed, you have

• `n = 1 implies {( n = 1 -> n=2), (n =1 -> n= 3), (n = 1 -> n=4), (n = 1 -> n=5) :}`

• `n=2 implies {(n = 2 -> n = 3), (n = 2 -> n= 4), (n = 2 -> n = 5) :}`

• `n = 3 implies {( n = 3 -> n = 4), (n = 3 -> n = 5):}`

• `n = 4 implies n = 4 -> n = 5`

This gives you a total number of transition equal to

`4 + 3 + 2 + 1 = 10`

This can be written as

`sum_(1)^(n -1) (n) = 1 + 2 + 3 + 4 = 10`

Similarly, when energy is released, you have

• `n= 5 implies {(n = 5 -> n = 4), (n = 5 implies n = 3), (n = 5 implies n= 2), (n = 5 -> n = 1) :}`

• `n = 4 implies {(n=4 -> n= 3), (n = 4 -> n = 2), (n = 4 -> n = 1) :}`

• `n = 3 implies {(n = 3 -> n = 2), (n = 3 -> n = 1) :}`

• `n = 2 implies n =2 -< n= 1`

Once again, the total number of transitions will be equal to

`4 + 3 + 2 + 1 = 10`

which can be written as

`sum_1^(n-1) (n) = 1 + 2 + 3 + 4 = 10`

If you add the transitions you get for absorption and the transitions you get for emission, you will have a total of `20` transitions.

You can say that the number of absorption transitions, which is equal to the number of emission transitions, can be found by using

`sum_1^(n-1)(n) = (n * (n-1))/2`

Therefore, you will have

`(n * (n-1))/2 = "no. of absorption OR emission transitions"`

`n * (n-1) = "no. of absorption AND emission transitions"`