Compute the Ka x ray energies of calcium (Z = 20), zirconium (Z = 40), and mercu

Question

Compute the Ka x ray energies of calcium (Z = 20), zirconium (Z = 40), and mercury (Z = 80). (Give your answers to at least three significant figures.) aimA zirconium mercury Xkev kev kev Compare with the measured values of 3.69 kev, 15.8 kev, and 70.8 kev. (Select all that apply.) The values computed are larger than the measured values. The values computed are smaller than the measured values. O The values computed are all within 5% of the measured values. The error of computed values decreases with larger z The error of computed values increases with larger Z.

Explanation / Answer

Solution:

We know that, for hydrogen like atoms, Ei = -13.6 ( Z2/ni2) ev

Calcium, Z= 20, for n = 2 to 1 (ni =2, nj =1)

Thus, E= -13.6 ( Z2/ni2) - [-13.6 ( Z2/nj2)] = 13.6 Z2 [ ( 1/nj2) - ( 1/ni2)] ev

==> E = 13.6 *(20)2 [ ( 1/12) - ( 1/22)] ev = 4080 ev = 4.08 keV (Answer) (For Calcium)

Compared to 3.69keV (measured value), percent difference = (4.08-3.69)*100/3.69 = 10.5% (larger)

Zirconium Z = 40 for n = 2 to 1 (ni =2, nj =1)

Thus, E= -13.6 ( Z2/ni2) - [-13.6 ( Z2/nj2)] = 13.6 Z2 [ ( 1/nj2) - ( 1/ni2)] ev

==> E = 13.6 *(40)2 [ ( 1/12) - ( 1/22)] ev = 16320 ev = 16.3 keV (Answer) (For Zirconium)

Compared to 15.8 keV (measured value), percent difference = (16.3 -15.8)*100/15.8 = 3.1% (larger)

Mercury Z = 80 for n = 2 to 1 (ni=2, nj=1)

Thus, E= -13.6 ( Z2/ni2) - [-13.6 ( Z2/nj2)] = 13.6 Z2 [ ( 1/nj2) - ( 1/ni2)] ev

==> E = 13.6 *(80)2 [ ( 1/12) - ( 1/22)] ev = 65280 ev = 65.3 keV (Answer) (For Mercury)

Compared to 70.8 keV (measured value), percent difference = (70.8 -65.3)*100/70.8 = 7.76% (smaller)

Option4: True Answer (The error of computed values decreases with larger Z)

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