Lead-206 is the end product of 238u decay. one 206pb atom has a mass of 205.974440 amu. (a) calculate the binding energy per nucleon in mev: mev/nucleon

The atomic number for Pb is 82

∴ Pb has 82 protons and 206-82 = 14 protons

The actual mass of Pb nuclei is

= (82 * mass of the proton) + (124 * mass of neutron)

= (82 * 1.00728) + (124 * 1.008664) amu

= 207.6713 amu

The mass of lead which is given is 205.9744 amu

∴mass defect is

m = 207.6713 - 205.9744 = 1.6969 amu

=1.6969 * 1.66054 * 10⁻²⁷kg

=2.818 * 10⁻²⁷kg

The binding energy is E = mc²

C is the speed of light in vacuum = 2.9979 * 10⁸m/s

∴ E = 2.532 * 10*⁻¹⁰ J/mol

= 2.532 * 10⁻¹⁰ * 6.023 * 10²³ J/mol

= 1.53811 * 10¹⁴ J/mol