Question

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. Assume that the coin is entirely made of \[_{29}^{63}Cu\] atoms (of mass 62.92960 u).

Options

• 2.5296 × 10²⁴ J

• 2.5096 × 10¹² J

• 2.6296 × 10²⁴ J

• 2.5296 × 10¹² J

Answer 1

2.5296 × 10¹² J

Explanation:

Mass of copper coin, m' = 3 g

Atomic mass of ₂₉Cu⁶³ atoms, m = 62.92960 u

The total number of ₂₉Cu⁶³ atoms in the coin,

\[N=\frac{N_A\times m^{\prime}}{\text{Mass number}}\]

where,

N = Avogadro's number

= 6.023 × 10²³ atoms/g

Mass number = 63 g

∴ \[N=\frac{6.023\times10^{23}\times3}{63}=2.868\times10^{22}\mathrm{atoms}\]

₂₉Cu⁶³ nucleus has 29 protons and (63 - 29) = 34 neutrons

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

Where,

Mass of a proton m_p = 1.007825 u

Mass of a proton m_n = 1.008665 u

So,

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

= 0.591935 u

Mass defect of all atoms present in the coin,

Δm = 0.591935 × 2.868 × 10²²

= 1.69766958 × 10²² u

But 1u = 931.5 MeV/c²

So, Δm = 1.69766958 × 10²² × 931.5 MeV/c²

So that, the binding energy of the nuclei of the coin is given as:

Ep = Δmc²

\[=1.69766958\times10^{22}\times931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right)\times c^{2}\]

= 1.581 × 10²⁵ MeV

But 1 MeV = 1.6 × 10^-13 J

E = 1.581 × 10²⁵ × 1.6 × 10^-13

= 2.5296 × 10¹² J

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