Question

18. Given that the mass of manganese-55 is 54.9380 u, calculate the total binding energy a. Write the nuclear reaction describing the formation of a manganese-55 nucleus. b. Calculate the change in mass in the reaction described in part a (mass of a neutron 1.0087 u; mass of a proton 1.0078 u). c. Calculate the binding energy (in joules) for one atom of manganese-55 d. Calculate the binding energy (in joules) for 1.000 g of manganese-55 10

Answer 1

a. 25 ₁H¹ + 30 ₀n¹ ----> ₂₅Mn⁵⁵

b. The change in mass ($\Delta$m) = (25*1.0078 + 30*1.0087) u - 54.9380 u

= 55.456 u - 54.9380 u

= 0.518 u

c. The binding energy for one atom of manganese ($\Delta$E) = $\Delta$mc²

= (0.518 u/nucleus * 1.6606*10^-27 kg/u) * (2.9979*10⁸ m/s)²

= 7.7309*10^-11 J/nucleus

d. 1 g of manganese = 1 g/54.9380 g.mol^-1 = 0.0182 mol = 0.0182 * 6.023*10²³ nuclei = 1.0963*10²² nuclei

Therefore, the binding energy for 1 g of manganese = 1.0963*10²² nuclei * 7.7309*10^-11 J/nucleus = 8.4756*10¹¹ J