What is the B(Z,A) of nucleus 250No -[Free] B83
What is the B(Z,A) of nucleus 250No
Question
What is the total binding energy B(Z, A) of the nucleus 250No (nobelium with mass number A = 250 and atomic number Z = 102)?
Answer
Detailed Explanation
To estimate the total nuclear binding energy, we use the semi-empirical mass formula:
B(Z, A) = avA − asA2/3 − ac × [Z(Z−1)] / A1/3 − aa × [(A−2Z)2 / A] + δ(A,Z)
For this calculation, the coefficients are:
- av = 15.8 MeV (volume term)
- as = 18.3 MeV (surface term)
- ac = 0.714 MeV (Coulomb term)
- aa = 23.2 MeV (asymmetry term)
Step-by-Step Calculation
1) Volume Term
15.8 × 250 = 3950 MeV
2) Surface Term
A2/3: (250)1/3 ≈ 6.30, then squared ≈ 39.7
18.3 × 39.7 = 726.5 MeV
3) Coulomb Term
Z(Z−1) = 102 × 101 = 10302
[Z(Z−1)] / A1/3 = 10302 / 6.30 ≈ 1636.2
0.714 × 1636.2 = 1169 MeV
4) Asymmetry Term
A−2Z = 250−204 = 46
(46)2 = 2116
2116 / 250 = 8.464
23.2 × 8.464 = 196.4 MeV
5) Pairing Term
Since both Z and N are even, δ = +12/A1/2
√A ≈ 15.81
δ ≈ +12/15.81 ≈ +0.76 MeV
Combining All Terms
B(Z,A) = 3950 − 726.5 − 1169 − 196.4 + 0.76
Step by step:
- 3950−726.5 = 3223.5
- 3223.5−1169 = 2054.5
- 2054.5−196.4 = 1858.1
- 1858.1 +0.76 = 1858.9 MeV
Final Result
The total binding energy of the nucleus 250No is approximately:
B(250,102) ≈ 1860 MeV
The binding energy per nucleon is:
≈ 1860 MeV / 250 ≈ 7.44 MeV/nucleon