Ncert Exercises & Solutions

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 up of \({ }_{29}^{63} \mathrm{Cu}\) atoms (mass of each atom = 62.92960 u).

Hint: The required energy will be equal to the total bindding energy of the nucleons.
Step 1: Fidn the number of atoms in the coin.

Mass of a copper coin, m = 3 gm

Atomic mass of

{}_{29}^{63}{Cu}

atomis in the coin, m = 62.92960 u

The total number of

{}_{29}^{63}{Cu}

atomis in the coin,

N = \frac{N_{A \times m'}}{mass Number}

where,

N_{A}

= Avogadro's number

= 6.023 \times 10^{23}

atoms/g and mass number = 63 g

N = \frac{6.022 \times 10^{23} \times 3}{63} = 2.868 \times 10^{22}

atoms
Step 2: Find the mass defect in \({ }_{29}^{63} \mathrm{Cu}\) nucleus.

{}_{29}^{63}{Cu}

nucleus has 29 protons and

(63 - 29)

34 neutrons.

∴

Mass defect of this nucleus,

∆ m = 29 \times m_{H} + 34 \times m_{n} - m

where, Mass of a proton,

∆

m= 1.007825u
Mass of a neutron,

m_{n}

=1.008665 u

∆

m= 29x1.007825+34x 1.008665-62.9296 = 0.591935 u
Step 3: Find the mass defect of the coin.

Mass defect of all the atoms present in the coin,

∆ M = 0.591935 \times 2.868 \times 10^{2} = 1.69766958 \times 10^{22} u

But 1 u = 931.5 MeV / c^{2}

∴ ∆ M = 1.69766958 \times 10^{22} \times 931.5 MeV / c^{2}

Step 4: Find the binding energy of the coin.
Hence, the binding energy of the nuclei of the coin is given as:

E_{b} = ∆M c^{2} = 1.69766958 \times 10^{22} \times 931.5 (\frac{MeV}{c^{2}}) c^{2} = 1.581 \times 10^{25} MeV

But 1 MeV = 1.6 \times 10^{- 13} J

E_{b} = 1.581 \times 10^{25} \times 1.6 \times 10^{- 13} = 2.5296 \times 10^{12} J

This much energy is required to separate all the neutrons and protons from the given coin.

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