Question

A Pt-204 atom (platinum-204) has a nucleus with 78 protons and an atomic mass of 203.980762 u. What is the nuclear binding energy per nucleon in this atom? 1.72 MeV O 1603 MeV 20.6 MeV O 12.7 MeV 7.86 MeV

Answer 1

Pt-204 $\rightarrow$ ₇₈Pt²⁰⁴

So, Pt-204 has 78 protons and (204-78)=126 neutrons.

Mass of 78 protons = 78 x 1.007276 u = 78.567528 u

Mass of 126 neutrons = 126 x 1.008665 u = 127.09179 u

So, total mass of 78 protons and 126 neutrons = 78.567528 + 127.09179 = 205.659318 u

Atomic mass of Pt-204 = 203.980762 u

So, mass defect, $\Delta$ M = 205.659318 - 203.980762 = 1.678556 u

$\because$ 1 u = 931.5 MeV/c²

So, binding energy, B = ($\Delta$M)c² =1.678556 x 931.5 = 1563.575 MeV

Number of nucleons, N = 204

So, binding energy per nucleon

                   $=\frac{B}{N}=\frac{1563.575}{204}=7.66\: MeV\approx \mathbf{7.86\: MeV}$