00:02
Hi, in this question, first we will consider the k alpha ray for which the transition is from n is equal to two state to n is equal to one state and the energy in the n state is given by e n is equal to minus 13 .6 into cid minus 1 square divided by n square.
00:28
That is the atomic number.
00:31
We are given with copper.
00:33
So for copper set is 29.
00:40
Now, first we'll consider n is equal to 2.
00:45
So e2 is equal to minus 13 .6 into 29 minus 1 of the whole square, divided by 2 square.
00:55
Upon solving this, we will get minus 266.
01:01
5 .6 ev and when n is equal to 1 e1 is equal to minus 13 .6 into 29 minus 1 the whole square divided by 1 square which is equal to minus 1066 2 .6 ev so we have the value for e 1 and e2 so energy for the k alpha ray that is e k alpha will be equal to e2 minus e1.
01:44
We know the value for e2 and e1.
01:47
So this will be equal to minus 2665.
01:52
0 .6 .6 minus minus of 10662 .6.
02:00
And upon solving this, we will get 7997...