# Question #2ab5b

##### 1 Answer

#### Answer:

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 *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 **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"#