Question

(a) Calculate the nuclide masses of H (Z=1, A=1), H (Z=1,A=3), He (Z=2,A=4), Fe (Z=26, A=56),Ni(Z=28,A=57),Au(Z=79,A= 197),U(Z=92,A=235) using the Weizsacker Formula (where A=Z+N) for the binding energy of the nucleon - Use the coefficients (Mayo) provided in class – which are the 2 most dominant terms and which is the least dominant?

(b) Compare and plot/tabulate the predicted values with published values of the isotope masses and explain if the formula can be utilized for reasonably accurate (i.e., < 1% error) predictions all over the atomic mass space

(c) In which “A” region does the formula start to give unacceptable inaccuracies?

Answer 1

Weizsacker Formula for Binding energy B(A,Z) in MeV is given by

22) B(AZ) = QA - B.A2/3 - 72 41/3-SA, Z)

Where A is Mass number and Z is atomic number

A = Z+N , where N is number of neutrons

In the above formula
a = 15.75 MeV
,
8 = 17.8MeV
,
7 = 23.7MeV
,

,

€ = 0.711 MeV

11.18 73

$\delta$ = 0 for odd-A, $\delta$=+1 for odd-odd nuclei and $\delta$=-1 for even-even nuclei

Following table lists the calculated Binding Energy value along with published results and % error

Element	A	Z	N	Calculated Binding Energy (MeV)	Calculated Binding Energy per nucleon(MeV)	Published value of Binding Energy per nucleon(MeV)	%error
H	1	1	0	-26.461	0	0	0
H	3	1	2	1.831528084	0.610509	2.827	78.5
He	4	2	2	22.35520289	5.588801	7.074	21
Fe	56	26	30	490.6614252	8.761811	8.79	0.3
Ni	57	28	29	488.8572616	8.576443	8.67	1.1
Au	197	79	118	1554.509533	7.890911	7.915	0.2
U	235	92	143	1785.898881	7.59957	7.59	0.1

For the region A $\geq$ 56 , agreement of Binding energy per nucleon is $\leq$ 1% error

Published values of Binding energy per nucleon for the above table is from https//periodictable.com