# Characteristic X-rays

## Homework Statement

The Kalpha x-ray is emitted when an electron undergoes a transition from the L shell (n = 2) to the K shell (n = 1). Use the following equations to calculate the wavelength of the Kalpha x-ray from a iron target (Z = 26), where E0 is the ground-state energy.

The problem is the number im using in my (Z - #) for the E_L shell. I am missing the concept here of how to know which number to use. If Someone can explain that part to me, this question will be answered

## Homework Equations

EK = −(Z − 1)2(13.6 eV)
EL = −Zeff2E2
λ =
hc
ΔE

## The Attempt at a Solution

E-k = -(26 - 1)^2(13.6 eV) = -8500 eV
E_l = -(26 - 4)^2(13.6 eV/4) = -1645.6 eV

λ = (4.14e-15 eVs)(3e8 m/s) / -1643.6 eV + 8500 eV = .181 nm

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Simon Bridge
Homework Helper
Use the following equations
[snip]
EK = −(Z − 1)2(13.6 eV)
EL = −Zeff2E2
What?

Is this:
$$E_K=-(Z-1)^2(13.6\text{eV}) \qquad E_L = -Z_{eff}^2E_2$$

For the the Bohr atom you are normally looking at:

$$E_n=-\frac{R_E (Z-1)^2}{n^2}: \qquad R_E=13.6eV$$
(you know the reduced Z is due to electron screening?)

The transition is from n=2 to n=1.

I know the Z is due to electrons shielding the nucleus. If the transition is from n=2 to n=1, then the amount of electrons screening should be just 1, considering it's an empty electron in the n=1 so there is only 1 shielding the rest. I took this into consideration and it still gave me the wrong answer. I saw another problem and they said it was (z - 4) when the problem involved nickel, I dont see how 4 electrons can possibly shield it if there can only be 2 electrons in the
k shell. You count the amount of electrons from one order below where the incoming electron is coming from, right?

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Simon Bridge
Homework Helper
That would be correct - the number of electrons below the transmission screens the charge. for Z-4 they may be taking sub-levels into account - for instance, if they are looking at the transition from the 2p1/2 (L2) state.

So what was the "correct" answer?
(Or is this one of those computer-mediated things? [1])

---------------------------
[1] if so - then try with the actual speed of light instead of 3e8m/s
note: a number good to memorize is: hc=1.2398 keV.nm

it is a computer generated question. I can't figure out why Z is not simply just 1. My answer is within 10% of the correct value it says. The only thing wrong is that # that needs to be there, do you have any idea what that # may be?\

Here is the solution when Nickel is the metal in question:
http://www.cramster.com/solution/solution/1084452

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Simon Bridge
Homework Helper
It should be Z-1. Moseley's law is E=(Z-1)2x10.2eV.

The computer will be sensitive to accumulated rounding and exact values.
I get an answer within 10% of yours by using the real c and #=1.
Give it a go.

I get transition energy of 6.375keV so the wavelength is 1.2398/6.375=0.194nm (3dp) which is 7% off from your result.

but you can clearly see in the solution that they DID NOT use Z - 1. They used Z - 4. Oh yeah you would need an account for it to work but

They Did this

E_L = -(28 -4)^2 (13.6 eV / 4) = -1958.4 eV

Why did you use 10.2 eV? It would be 13.6 eV since that is the ionization of hydrogen

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Simon Bridge
Homework Helper
I cannot see anything in the solution - it won't show me.
I cannot even see the question the solution refers to.

Have you even tried the suggestion?

Note: the L-shell has three energy levels, one 2s, and two 2p.
The electron in the 2s state has the nucleus screened by 1, in the lower 2p it is screened by 3 and in the upper 2p is screened by 4.
[edit: can't flaming count! in the lower 2p it is 4: 1 in the 1s, 2 in the 2s and the other one in the 2p]
So that is their reasoning.

The transition is usually considered from the lowest state unless the transition from the edge is specified.
Was it specified? Of course, you could always try all three and see which one gets accepted.

Basically, afaict you are doing it right. The only reason for the computer to reject you is because the numbers don't quite match - i.e. if you have rounding errors, or you have approximated something in a way the program has not anticipated. I never use computer-mediated homework for exactly this reason. Unless your course wants you to use a formula I don't know about.

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Yes sir, and it was wrong. The question is identical except instead of being iron with Z = 26, it's nickel with Z = 28.

I GOT IT! However, I used (Z - 3) to get it. Somehow, 3 was the right number. Any insight? It was .185 nm

Simon Bridge
Homework Helper
You'll have to look at the fine detail of the electronic structure of iron :(
Could I have been right the first time and the lower 2p state is only screened by 3?

Simon Bridge
Homework Helper
Experimentally, the K-alpha transition energies are
6.391keV and 6.404keV

This yields 0.1939nm and 0.1935nm - so the first is the 2s > 1s transition and the second is the 2p > 1s transition (big jump = small wavelength).

I got 0.194nm

You see that mine gives the closer approximation - which is why it is unusual to use lots of screening. Basically, the only difference between the whole atom and the ionized one is that missing electron in the 1s state - so you only need to adjust the Bohr equation by 1 to get a good approximation.

Someone may want to give this a go for Tin.

The reasonable conclusion is that adjusting the L shell energy level for the screening effect of the K shell electrons does not provide a good approximation for measured energy transitions. This should makes sense since otherwise the L shell energy would need to be adjusted even more when the K shell is full.

Hopefully your course will point this out later.

aside: I want to correct #8 counting.
The K shell has one subshell - 1s
The L shell has two - 2s and 2p
Each sub-shell can have 2 electrons in it.
The ionized atom has one less electron in the 1s state.

if the transition is 2s - 1s, there is one screening electron
if the transition is 2p - 1s, there are three

however, one 2p electron has a higher energy than the other one ... so the maximum jump is from a state with four screening electrons.

This is why we have seen the Z-1, Z-3, and Z-4 adjustments above.