A given coin has a mass of 3.0g. What is 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 `""_(29)^(63)Cu` atoms (of mass 62.9260 u).

Options

A. `1.58xx10^(25)` MeV
B. `1.23xx10^(19)` MeV
C. `1.26xx10^(16)` MeV
D. `1.26xx10^(16)` MeV

Answer

A

Explanation

Given, mass of coin=3g, atomic mass of Cu=63
mass of `._(29)^(63)Cu,M=69.92960`,
Avogadro's number=`6.023xx10^(23)`
`m_(p)=1.007725u and m_(n)=1.008665u`
Nuclear energy reuired to required to separte neutrons and protons, `E_(b)=?`
Each atom of copper contains 29 protons and 34 neutrons. Therefore, calculate the mass defect the mass defect of each atom using the relation,
`DeltaM=[Zm_(p)+(A-Z)m_(n)]-M`
`DeltaM=[29xx1.007825+34xx1.008665]-62.92960`
`=0.591935u`
Number of atoms in 3 g coin`=(6.023xx10^(23)xx3)/(63)=2.868xx10^(22)`
Total mass defect of all atoms,
`(Deltam)_("total")=0.591935xx2.868xx10^(22)=1.6977xx10^(22)`
The nuclear energy required `E_(b)` to separte all the neutrons and protons from each other, by using the relation,
`E_(b)=(Deltam)xxc^(2)=(Deltam)c^(2)xx931MeV//c^(2)" "[because1u=931MeV]`
`=1.58xx10^(25)` MeV