# Finding wavelength of hydrogen atom when changing energy states

1. May 22, 2010

### kavamo

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

Find the wavelength of the radiation emitted when a hydrogen atom makes a transition from the n = 6 to the n = 3 state. Give answer in Âµm.

2. Relevant equations

z^2 / n^2 x 13.6ev
delta E = E2 - E1 = hv

3. The attempt at a solution

E(sub n1) = 1^2 / 6^2 x 13.6 ev = 0.3778
E(sub n2) = 1^2 / 3^2 x 13.6 ev = 1.511

1.511 - 0.3778 = 1.1332 = hv

1.1332 / 6.626 x 10^-34 = v

v = 1.710 x 10^33

I am very confused by this problem and am not even sure if I have the right equations, which were taken from the book. Please advise. Thank you in advance.

2. May 22, 2010

### Redbelly98

Staff Emeritus
You are doing good up to this point, except that you are not showing what the units are in this energy difference calculation. So, what are the units that go with the numbers 0.3778, 1.511, and 1.1332?

Likewise, what units go with the 6.626 x 10^-34 value? Include the units with 1.1332 and 6.626 x 10^-34, and that will get to what is wrong with your calculation.[/quote]
You do have the correct equations, and the right approach to solving it. The key is in the units.

3. May 22, 2010

### kavamo

I think that all the units are in ev's but am not sure as they are not given. You have found the part of this thatis the hardest for me--units.

4. May 22, 2010

### kavamo

I found the minimum photon energy required to study this molecule is 8.378 kev. will i use this in the above problem?

5. May 22, 2010

### Redbelly98

Staff Emeritus
That's correct for the 0.3778, 1.511, and 1.1332 values. Next look up Planck's constant h, it should be in your textbook. That will give you the units that go with 6.626 x 10^-34.

That is not relevant here.

6. May 22, 2010

### kavamo

Planks constant units are J/s

7. May 22, 2010

### Redbelly98

Staff Emeritus
Actually it's J s. (Multiply, don't divide, the s).
Convert that to eV s, and then things should work out.

8. May 22, 2010

### kavamo

I'm still a little confused. Am I doing this correctly?

v = 1.710 x 10^33 eV =hc / E delta E = 1.1332 eV

v = (6.626 x 10^-34) Js x (3 x 10^8)m/s divided by 1.1332eV

= 1.754E-25

9. May 22, 2010

### Redbelly98

Staff Emeritus
That doesn't look right at all.

Can you convert the 1.13 eV into J? Then use ΔE=hv as before. You need to have ΔE and h with the same energy units -- it could be eV or J, but it has to be the same for both.

10. May 22, 2010

### kavamo

Thank you for your help. I was able to get help from a classmate in between posts and she helped me work it through. Thanks again.