Absorption and Emission in the Hydrogen Atom?

Question

A monochromatic laser is exciting hydrogen atoms from the n = 2 state to the n = 5 state.

Wavelength of laser = 435nm

1) Eventally all of the excited hydrogen atoms will emit photons until they fall back to the ground state. How many different wavelenghts can be observed in this process? (Answer not 3)

2) What is the longest wavelength that is observed? (Answer not 2.28*10^-6)

3) What is the shortest wavelenght observed? (Ans not 3.65*10^0-7)

Answer 1

1)

Firstly looking at the energy level plot you should notice that n=1 is defined as the ground state. So if they are dropping to the ground state then they can make any order of jumps from 5 to 1. This means the following order of jumps.

5 - 4 - 3 - 2 - 1

5 - 4 - 2 - 1

5 - 4 - 3 - 1

5 - 4 - 1

5 - 3 - 2 - 1

5 - 3 - 1

5 - 2 - 1

5 - 1

Check to make sure i haven't missed any in there but i am sure you get the idea.

EDIT: Answer should be 10 because in the sequence I wrote you have 5 to 4, 5 to 3, 5 to 2, 5 to 1, etc. to 2 to 1

2)

If you look at the specific energy values for the various electron excitation levels you have:

n = 1     -13.6ev

n = 2     -3.4ev

n = 3     -1.511ev

n = 4     -0.850ev

n = 5     -0.544ev

Now the longest wavelength corresponds to the smallest energy release.Specifically this is from -0.544 to -0.850. The difference has a magnitude of 0.306eV and we can calculate the wavelength as:

lambda = c * h / E

c - speed of light (m/s)

h - planck's constant

E - energy difference between valence levels

lambda = 4.05E-6

3)

I you want to know the shortest wavelength you would see that this corresponds to the largest valence energy difference which is simply from n=5 to n=1. This difference is 13.056eV.

Using the same equation as in 2 you get:

lambda = 9.50E-8

Hopefully this helps you out some. Read through this carefully and confirm it.