Using Moseley's Law to Calculate Kalpha2 for Lead with Z=82

[mss90]

Homework Statement

I need to use Mosley law to find Kalpha2 for lead where Z=82

What i know is that copper (z=29) and tin (z=50) have kalpha2 of 8.03 and 25 KeV respectively

Homework Equations

Mosley law

The Attempt at a Solution

Could someone give me a hint please

 

[NascentOxygen]

Try a google search, using the correct spelling, Moseley's Law. You had it correct in the title line.

You'll find the formula here, and it checks out with the data you provided. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/moseley.html

(Disclaimer. I have no idea how realistic it is to apply that formula to values of Z as high as 82.)

 

[mss90]

just use; sqrt{f} = k_1 * (Z - k_2)

f=E/h then add inn the constant and the energy but what i don't understand is how knowing those two (for Cu and Sn) can help with finding the third?

Also where can i acquire the constants? It just states;

and

are constants that depend on the type of line

thanks

 

[NascentOxygen]

I think it is all in the article I linked to. Perhaps you didn't scroll down?

By substituting Z for Cu and Sn you can confirm that the equation works and gives the energy in eV just as you require.

 

[mss90]

okay so k1=2.47*10^15 and k2 = 1 for kalpha?
How does knowing the energies for these 2 other elements help me figure it out? I am not getting this :P

Also, the question is talking about Kapha2 not just Kalpha, does that matter?

 

[NascentOxygen]

Could you please post a jpeg of the question as it's been given to you.

 

[mss90]

 

[NascentOxygen]

Okay. Here's my take on this ... and I could be wrong.

I think Moseley's Law equation is all that's called for. The difference in transition energies K_α1 and K_α2 is tiny and it suffices to use Moseley's formula as for K_α. (For an explanation of the doublets, refer to the diagram here: http://en.m.wikipedia.org/wiki/K-alpha )

Maybe the examiners didn't expect students to have memorised the constants in the formula, just the bare relationship
E = k(Z - 1)²

So, by providing you with a couple of its results, you can determine k on the spot without having memorised it. Finally, using the value of k you calculated or had memorised apply Moseley's equation for atomic number 82 and determine K_α.

Sound plausible?

 

[NascentOxygen]

I came across a table listing K_a1 and K_a2 which look suspiciously like they may be K_α1 and K_α2 though I can't see that stated. If I'm right, you can compare them to see how close their values are. http://hyperphysics.phy-astr.gsu.edu/hbase/tables/kxray.html#c1

 

[mss90]

E=k(Z-1)2 ó k=(Z-1)2/E
(29-1)2/8.03=97.63

(50-1)2/25 =96.04

theyr pretty close but not identical so could i just the average for K to figure out the energy for lead?

 

[mss90]

ye i saw that table as well, thanks

 

[NascentOxygen]

E=k(Z-1)2 ó k=(Z-1)2/E
(29-1)2/8.03=97.63

(50-1)2/25 =96.04

theyr pretty close but not identical so could i just the average for K to figure out the energy for lead?

That sounds like a good experimental approach.

What year of the physics course is this test paper?

 

[mss90]

E=k(Z-1)2 hence E=96.835(82-1)2 = 635334.435J its pretty high right? I am guessing its in joules
when converting I am getting 3.9654458121e+24eV doesn't compare with the table

 

[NascentOxygen]

Your post #1 indicates your working is in keV!

 

[mss90]

Your post #1 indicates your working is in keV!

Ye but it still doesn't line up, right?

 

[NascentOxygen]

E=k(Z-1)2 ó k=(Z-1)2/E

You haven't rearranged this correctly to isolate k.

 

[mss90]

aaah that's embarrassing, its E on top :p
E=k(Z-1)2 ó k=E/(Z-1)2
8.03 / (29-1)2=0.0102
25 / (50-1)2 =0.0104

Average = 0.0102+0.0104/2=0.0103
E=k(Z-1)2 hence E=0.0103 *(82-1)2 = 64.32KeV