Finding wavelength of hydrogen atom when changing energy states

[kavamo]

Homework Statement

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 µm.

Homework Equations

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

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.

 

[Redbelly98]

Homework Statement

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 µm.

Homework Equations

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

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

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?

1.1332 / 6.626 x 10^-34 = v

v = 1.710 x 10^33

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]

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.

You do have the correct equations, and the right approach to solving it. The key is in the units.

 

[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.

 

[kavamo]

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

 

[Redbelly98]

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.

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.

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

That is not relevant here.

 

[kavamo]

Planks constant units are J/s

 

[Redbelly98]

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

 

[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

 

[Redbelly98]

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.

 

[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.