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

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..
 

Dick

Science Advisor
Homework Helper
26,249
611
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.
 

nrqed

Science Advisor
Homework Helper
Gold Member
3,528
178
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.
 
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.
 

nrqed

Science Advisor
Homework Helper
Gold Member
3,528
178
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!
 
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
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top