Need help calculating wavelengths of L(alpha) and L(beta

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In summary, The conversation is about a question regarding the calculation of the wavelength of L(alpha) and L(beta) x-ray photons produced from a copper target (Z=29). The approach used is E=hc/lambda and rearranging for lambda, but there is confusion about the speed of alpha and beta rays and the specific electron transitions being referred to. The poster plans to clarify the question with their tutor and update the thread once it is solved.
  • #1
obi-wan
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Hi, I need some guidance on a particular question for my coursework.

The questions states: Calculate the wavelength of L(alpha) and L(beta) x-ray photons produced from a copper target (Z=29)


I have tried finding the wavelength using E=hc/lambda and re-arranged for lambda (lambda=hc/E)

Is this the correct approach to solving the problem? Alpha and beta rays travel at the same speed ~ 3*10^8m/s, seems like I'm missing something.

I don't want answers as this doesn't help me in any way but just need some guidance as to how it can be worked out.

Your help is much appreciated..
 
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  • #2
Alpha and beta rays no not travel at the same speed, they aren't photons. They are likely talking about specific electron transitions in copper. I'm guessing they are talking about Lyman-alpha and Lyman-beta. Look those up and see if it looks like that's what L(alpha) and L(beta) mean.
 
  • #3
obi-wan said:
Hi, I need some guidance on a particular question for my coursework.

The questions states: Calculate the wavelength of L(alpha) and L(beta) x-ray photons produced from a copper target (Z=29)


I have tried finding the wavelength using E=hc/lambda and re-arranged for lambda (lambda=hc/E)

Is this the correct approach to solving the problem? Alpha and beta rays travel at the same speed ~ 3*10^8m/s, seems like I'm missing something.

I don't want answers as this doesn't help me in any way but just need some guidance as to how it can be worked out.

Your help is much appreciated..


I also certain they are talking about Lyman alpha and beta transitions, like Dick suggested. The only problem is that copper has obviously many electrons.

So maybe they are referring to transitions of the last electron (the first 28 electrons fill the n=1,2 and 3 shells). So maybe they are referring to the transition between n=4 and n=5? I am not sure what they would use for the charge of the screened nucleus. The most simple-minded thing to try is Z=1 but in real life the outer electron would see a screened value a bit different than this.

On the other hand, Lyman usually refers to transitions from n=1 so would they assume a hydrogen-like copper atom? (one for which 28 electrons have been stripped off?) In which case one would use the Bohr energy level equation with Z=29. But this interpretation seems unlikely.
 
  • #4
When x-rays strike an electron on a copper surface this produces a radiation and a recoiling electron. However this radiation is not in the form of gamma or beta rays (correct me if I'm wrong here)

The question seems to be missing something and I can't think of anything else that would relate the gamma and beta rays to x-ray photons. The question does not talk about transitions between any energy levels.. best to ask my tutor to see if he makes sense of it.

thanks for your time Dick & nrqed.
 
  • #5
obi-wan said:
When x-rays strike an electron on a copper surface this produces a radiation and a recoiling electron. However this radiation is not in the form of gamma or beta rays (correct me if I'm wrong here)
I don't understand the question then. The question askes about X ray photons produced by Cu atoms. Here you are talking about bombarding Cu atoms with x-rays. I did not think that was the question. But I am maybe completely misinterpreting.

Better clarify things with your tutor.

best luck!
 
  • #6
sorry I've confused you there :) yes the question does ask about x-ray photons being produced by a Cu target. Nevermind, I will see my tutor tomorrow and get him to clarify things. Will keep this thread updated once the question makes sense, hopefully solving it and marking the thread as 'Solved'

thank you nrqed
 

What is the formula for calculating the wavelength of L(alpha)?

The formula for calculating the wavelength of L(alpha) is: wavelength = (h*c)/E, where h is Planck's constant, c is the speed of light, and E is the energy of the photon.

How do I determine the energy of the L(alpha) photon?

The energy of the L(alpha) photon can be determined using the Rydberg formula: E = -Rhc/(n^2), where R is the Rydberg constant, h is Planck's constant, c is the speed of light, and n is the principal quantum number.

What are the values of h and c used in the calculation?

The value of Planck's constant (h) is 6.626 x 10^-34 Joule seconds, and the speed of light (c) is 299,792,458 meters per second.

Do I need to convert the energy to electron volts (eV) before calculating the wavelength?

Yes, the energy should be converted to electron volts (eV) before using it in the wavelength formula. 1 eV is equal to 1.602 x 10^-19 Joules.

How do I calculate the wavelength of L(beta)?

The wavelength of L(beta) can be calculated using the same formula as L(alpha), but with a different energy value. The energy of L(beta) is E = -Rhc/((n+1)^2), where n is the principal quantum number.

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