Question:

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 2963Cu^{63}_{ 29}Cu atoms (of mass 62.92960 u).

Updated On: Sep 29, 2023
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Solution and Explanation

Mass of a copper coin, m=3gm’ = 3 g
Atomic mass of 2963Cu^{63}_{ 29}Cu atom, m = 62.92960 u
The total number of 2963Cu^{63}_{ 29}Cu atoms in the coin
N = NA× mMassNumber\frac{NA \times  m'}{Mass Number}
Where, 
NA = Avogadro’s number = 6.023 × 1023 atoms /g 
Mass number = 63 g
N=6.023× 1023× 363=2.868× 1022atomsN =\frac{6.023 \times  10^{23} \times  3 }{63} = 2.868 \times  10^{22} atoms
2963Cu^{63}_{ 29}Cu nucleus has 29 protons and (63 − 29) 34 neutrons
Mass defect of this nucleus, ∆m' = 29 × mH + 34 × mn − m 
Where, 
Mass of a proton, mH = 1.007825 u 
Mass of a neutron, mn = 1.008665 u 
∆m' = 29 × 1.007825 + 34 × 1.008665 − 62.9296 
= 0.591935 u 
Mass defect of all the atoms present in the coin, 
∆m = 0.591935 × 2.868 × 1022 = 1.69766958 × 1022
But 1 u = 931.5 MeVc2\frac{MeV}{c^2}
∆m = 1.69766958 × 1022 × 931.5 MeVc2\frac{MeV}{c^2}
Hence, the binding energy of the nuclei of the coin is given as: 
Eb=mc2E_b= ∆mc^2
= 1.69766958 × 1022 × 931.5 (MeVc2)c2(\frac{MeV}{c^2})c^2
= 1.581 × 1025 MeV 
But 1 MeV = 1.6 × 10−13
Eb = 1.581 × 1025 × 1.6 × 10−13 = 2.5296 × 1012
This much energy is required to separate all the neutrons and protons from the given coin.

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