# 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. - Physics

Numerical

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"Cu" atoms (of mass 62.92960 u).

#### Solution

Mass of a copper coin, m' = 3 g

Atomic mass of ""_29^63"Cu" atom, m = 62.92960 u

The total number ""_29^63"Cu" of atoms in the coin, "N" = ("N"_"A" xx "m'")/"Mass number"

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms/g

Mass number = 63 g

∴ N = (6.023 xx 10^23 xx 3)/63

= 2.868 × 1022 atoms/g

""_29^63"Cu" nucleus has 29 protons and (63 − 29) 34 neutrons

∴ Mass defect of this nucleus, Δm' = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∴ Δm' = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022

= 1.69766958 × 1022 u

But 1 u = 931.5 MeV/c2

∴ Δm = 1.69766958 × 1022 × 931.5 MeV/c2

Hence, the binding energy of the nuclei of the coin is given as:

Eb = Δmc2

= 1.69766958 × 1022 × 931.5 ("MeV"/"c"^2)xx "c"^2

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 1.581 × 1025 × 1.6 × 10−13

= 2.5296 × 1012 J

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

Concept: Nuclear Energy - Introduction of Nuclear Energy
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.5 | Page 462
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 5 | Page 462
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