A given coin has a mass of $3.0 \mathrm{~g}$. 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 ${ }_{29}^{63} \mathrm{Cu}$ atoms (of mass $62.92960 \mathrm{u}$ ).
Mass of a copper coin, m’ = 3 g
Atomic mass of ${ }_{29} \mathrm{Cu}^{63}$ atom, $m=62.92960 \mathrm{u}$
The total number of ${ }_{29} \mathrm{Cu}^{63}$ atoms in the coin, $N=\frac{N_{\mathrm{A}} \times m^{\prime}}{\text { Mass number }}$
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
$\therefore N=\frac{6.023 \times 10^{23} \times 3}{63}=2.868 \times 10^{22}$ atoms
${ }_{29} \mathrm{Cu}^{63}$ 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:
$E_{b}=\Delta m c^{2}$
$=1.69766958 \times 10^{22} \times 931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right) \times c^{2}$
= 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.
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- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
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- Current Electricity
- Capacitor
- Magnetism and Matter
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- Dual Nature of Matter
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- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
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