Answer
(a) $E_B = 76.21 MeV $
(b) $E_B = 76.68 MeV$
$Percentage = 0.62 \%$
The equation shows greater percentage accuracy for $^{62} Ni$
Work Step by Step
$(a)$ The binding energy of $_{\space \space 5}^{11}B$ is
$E_B = (Zm_H + (A-Z)m_n -m_{B-11})c^2$
$E_B = ((5)(1.007825 u) + (11-5)(1.008665 u) - 11.009305 u)931.5 MeV/u$
$E_B = 76.21 MeV $
$(b) $ another way to calculate binding energy is
$E_B = C_1A - C_2A^{2/3} - C_3\frac{Z(Z-1)}{A^{1/3}} +C_4\frac{(A-2Z)^2}{A} \pm C_5A^{-4/3} $
Where
C1 = 15.75 MeV
C2 = 17.80 MeV
C3 = 0.7100 MeV
C4 = 23.69 MeV
C5 = 39 MeV
$E_B = (15.75 MeV)(11) - (17.80 MeV)(11)^{2/3} - (0.7100 MeV)\frac{5(5-1)}{11^{1/3}} +(23.69 MeV)\frac{(11-2(5))^2}{11} \pm 0 $
* Note that $C_5 = 0$ because the number of Z is odd and the number of N is even. So,
$E_B = 173.25 MeV - 88.04 MeV - 6.385 MeV - 2.15 MeV$
$E_B = 76.68 MeV$
Percentage difference can be calculated by
$Percentage = \frac{76.78 MeV- 76.21 MeV}{76.21} \times 100 $
$Percentage = 0.62 \%$
Eq. (43.11) has a greater percentage accuracy for $^{62} Ni$ because this semi-empirical mass formula is more accurate for heavier nuclei.
