# Homework Help: Kinetic Enegry and Hydrogen atoms

1. Nov 9, 2008

### lebprince

1. The problem statement, all variables and given/known data

a) What is the minimum kinetic energy in electron volts that an electron must have to be able to ionize a hydrogen atom (that is, remove the electron from being bound to the proton)? I got the answer here which is 13.6 ev

b) If electrons of energy 12.8 eV are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?
b-1)Energy of highest-energy photon: I got the answer here which is 12.8
b-2)Energy of next highest-energy photon:
b-3)Energy of next highest-energy photon:
b-4)Energy of next highest-energy photon:
b-5)Energy of next highest-energy photon:
b-6)Energy of lowest-energy photon:

I Just need help with ( b-2 to b-6) Thanks

2. Nov 9, 2008

### Redbelly98

Staff Emeritus
The answer depends on details of your solution to b-1. In particular, what energy level (n value) is the Hydrogen excited to by the 12.8 eV energy?

3. Nov 9, 2008

### lebprince

am sorry but that 12.8 was a guess for b-1 and i dont know how to find that.

4. Nov 9, 2008

### borgwal

Question (b) is a bad question: the energies of the photons emitted by an actual gas form a continuum because of the motion of the atoms, their collisions, and the fact the excited states are unstable. So you can't even talk about the energy of the "next highest-energy" photon.

But setting that aside, what is meant here, no doubt, is that the 12.8 eV brings you to a particular excited state with a certain value of n, call it n_0 (easy to calculate). From n_0 the atom could make a transition to the ground state, giving back a 12.8 eV photon. But it could also make a transition to the n=2 state, emitting a less energetic photon. It could also "fall" to the n=3 state, emitting an even less energetic photon. And, once an atom is in the n=3 state, it could fall to the ground state from there, emitting yet another photon....etc.

5. Nov 9, 2008

### lebprince

so is there a specific formula i should use to calculate these answers?

6. Nov 9, 2008

### borgwal

Yes...don't you know a formula for the energies of hydrogen levels, as a function of "n"?

The question clearly assumes you know this, so look it up in your textbook!

7. Nov 9, 2008

### lebprince

the formula i know is -13.6/n^2 oh so instead of 13.6 i would use 12.8?

8. Nov 9, 2008

### borgwal

No, it is and remains -13.6eV/n^2.

9. Nov 9, 2008

### borgwal

Hint: the ground state corresponds to n=1. How much energy difference is there between n=1 and n=2?

You want to find a n_0 such that the difference between the levels n_0 and n=1, is about 12.8 eV.

10. Nov 9, 2008

### lebprince

so n0 - n1 = 12.8 and n1 is -13.6 so n0= 12.8-13.6 = -0.8

11. Nov 9, 2008

### borgwal

Do you understand the equation that says the energy of level n is -13.6eV/n^2?

12. Nov 9, 2008

### lebprince

it doesnt look like it cause the teacher we have is horrible. i barely understand the question sorry.

13. Nov 9, 2008

### borgwal

It says the energy of level n=1 is -13.6 eV, for n=2 the energy is -13.6/4=-3.4 eV, etc.
So the difference in energy between these two levels is (-3.4--13.6)eV=10.2 eV.

Now find n_0 such that -13.6/n_0^2+13.6=12.8 eV.

14. Nov 9, 2008

### lebprince

ok so n0^2= 13.6/0.8 = 17 then sqrt 17 = 4.12 which n0

15. Nov 9, 2008

### borgwal

Yes, so presumably n=4 is meant as the level in which the hydrogen atoms end up (n can only be an integer!)

16. Nov 9, 2008

### lebprince

ok good so for b-2 i would use n= 3 so -13.6/9+13.6 and that gave me 12.089
for b-3 n=2 so i would get 10
b-4 i would use n=1 so now i would get 0 which gave me a wrong answer the first 2 were right how come

17. Nov 9, 2008

### borgwal

You can go from n=4 to n=2 also

18. Nov 9, 2008

### lebprince

so i should get the same 3 answer again? if so its not working.

19. Nov 9, 2008