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Q.13.5 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 _{29}^{63}\textrm{Cu}  atoms (of mass 62.92960\; \; u).

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Mass of the coin is w = 3g

Total number of Cu atoms in the coin is n

 \\n=\frac{w\times N_{A}}{Atomic\ Mass}\\ n=\frac{3\times 6.023\times 10^{23}}{62.92960}

n=2.871\times1022

mH = 1.007825 u

mn = 1.008665 u

Atomic mass of _{29}^{63}\textrm{Cu} is m=62.92960 u

Mass defect \Deltam=(63-29)\timesmn+29\timesm- m

\Deltam=34\times1.008665+29\times1.007825 - 62.92960

\Deltam=0.591935 u

Now 1u is equivalent to 931.5 MeV

Eb=0.591935\times931.5

Eb=551.38745 MeV

Therefore binding energy of a _{29}^{63}\textrm{Cu} nucleus is 551.38745 MeV.

The nuclear energy that would be required to separate all the neutrons and protons from each other is

n\timesEb=2.871\times1022\times551.38745

=1.5832\times1025 MeV

=1.5832\times1025\times1.6\times10-19\times106J

=2.5331\times109 kJ

 

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Sayak

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