Atomic Physics - Xrays and energy

[pat666]

Homework Statement

5 The energy levels of a molybdenum atom are approximately:
K shell ….. – 20.00 KeV
L shell …... – 2.52 KeV
M shell …... – 0.23 KeV

i) Estimate the energy of the K_α, K_$$\beta$$ and L_α x-ray emissions.
ii) What are the wavelengths of these emissions?

Homework Equations

The Attempt at a Solution

I think K_$$\alpha$$ is from L shell to k shell??
so E_K$$\alpha$$=-2.52KeV--20KeV
E=17.48KeV

and the K_$$\beta$$ is from M shell to K shell?
then find the difference again for the energy

I don't know what the L_$$\alpha$$ is but possibly L shell to ground state or M shell to L shell??

not sure if anything I am doing there is right?

ii)
shouldn't be hard once I have energy's
E=hf so E=hc/$$\lambda$$

not sure if anything I am doing here is right?

Thanks for any help!

 

[ehild]

Read this: http://en.wikipedia.org/wiki/X-ray_fluorescence

ehild

 

[pat666]

Thanks ehild and wiki,
so the Lα transition is from M to L.
I'm a bit confused as to whether or not the element is relevant to any of the calculations, from what I read on wiki it seems related by the amount of energy gained form each transition but since I have those values I shouldn't even need to know that the element is molybdendum?
also I did the wave length calculation for the first transition and it came out to 1.24*10^-9m, this seems about right for an xray? (changed energy from KeV to J)

Thanks

 

[ehild]

The x-ray lines are named by the letter of the final state. The element is not relevant as you were given the level energies. Check the magnitude of the wavelength, I got a different result.

ehild

 

[pat666]

Yeah, I converted 17.48eV instead of 17.48KeV
now I get 7.1*10^-11 m?
seems like a tiny wavelength.
thanks.

 

[ehild]

X rays are of very short wavelength. Read the first sentences of the article. ehild

 

[pat666]

Thanks for helping ehild!