Question

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

Answer 1

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