Exercise 5: Labeling Label the following illustrations. A....

Transcript

00:01 Hello students, so for this video we have some electronic transitions and we have to find the energy, lowest energy photon released.

00:08 So we know that energy of a photon will be directly proportional to the energy difference between initial and final state of an atom.

00:32 That means that if the energy difference, the energy difference which is the least will have the least energy photon released.

00:41 So the first transition is from n is equal to 5 to n is equal to 1.

00:49 So the formula that we are going to use is delta e is equal to minus 13 .6 z square 1 by nf square minus 1 by ni square.

01:03 So nf means the final state and ni means the initial state where the electron was.

01:08 And since we are doing it for hydrogen, so z equals to 1.

01:11 So we will omit it in this calculations.

01:15 So for the first transition, the delta e is equal to minus 13 .6 into 1 by 1 square because the final is 1 by 1 by 5 square.

01:28 And this gives the value as minus 13 .056.

01:33 Now for the second transition, it is from n is equal to 6 to n is equal to 5.

01:42 So delta e is equal to minus 13 .6.

01:47 So 1 by 6 to 5.

01:50 So 5 square minus 1 by 6 square.

01:52 That is equal to minus 13 .6.

01:55 1 by 25 minus 1 by 36.

01:58 And this gives the value as minus 0 .166.

02:04 Now for the third one, we have from n is equal to 2 to n is equal to 5.

02:12 So it is jumping.

02:13 So delta e is equal to minus 13 .6.

02:18 1 by final minus initial...

Exercise 5: Labeling Label the following illustrations. A. Primary, exit, and attenuated photons. 6. No interaction 5. Primary 1 2 3 4 Primary -- Exit = Attenuation 1. 2. 3. 4.