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 6329Cu atoms (of mass 62.92960 u).

Answer 1

answer : 1.6 × 10^25 MeV

Let's first find binding energy of each copper nucleus and then we can easily find binding energy of 3.0 g of $^{63}_{29}Cu$.

we know,

• mass of each proton = 1.00783u
• mass of each neutron = 1.00867u

in $^{63}_{29}Cu$ presents 29 protons and (63 - 29) = 34 neutrons.

now mass of 29 protons = 29 × 1.00783 = 29.22707u

mass of 34 neutrons = 34 × 1.00867u

= 34.29478u

Total theoretical mass = 29.22707u + 34.29478u = 63.52185u

mass defect , ∆m = 63.52185u - 62.92960u = 0.59225u

so, binding energy of each Cu nucleus = ∆m × 931MeV

= 0.59225 × 931 = 551.385MeV

let's find number of cu atoms in 3g .

number of cu atoms = 3g/63g × 6.023 × 10²³ = 2.86 × 10²²

Total binding energy in 3g of copper = 2.86 × 10²² × 551.385 MeV = 1.57696 × 10^25 MeV ≈ 1.6 × 10^25MeV

= 1.6 × 10^25 MeV

Answer 2

Explanation:

answer : 1.6 × 10^25 MeV

Let's first find binding energy of each copper nucleus and then we can easily find binding energy of 3.0 g of ^{63}_{29}Cu

29

63

Cu .

we know,

mass of each proton = 1.00783u

mass of each neutron = 1.00867u

in ^{63}_{29}Cu

29

63

Cu presents 29 protons and (63 - 29) = 34 neutrons.

now mass of 29 protons = 29 × 1.00783 = 29.22707u

mass of 34 neutrons = 34 × 1.00867u

= 34.29478u

Total theoretical mass = 29.22707u + 34.29478u = 63.52185u

mass defect , ∆m = 63.52185u - 62.92960u = 0.59225u

so, binding energy of each Cu nucleus = ∆m × 931MeV

= 0.59225 × 931 = 551.385MeV

let's find number of cu atoms in 3g .

number of cu atoms = 3g/63g × 6.023 × 10²³ = 2.86 × 10²²

Total binding energy in 3g of copper = 2.86 × 10²² × 551.385 MeV = 1.57696 × 10^25 MeV ≈ 1.6 × 10^25MeV