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. For simplicity assume that the coin is entirely made of `""_29^63"Cu"` atoms (of mass 62.92960 u).

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).
Given m_p = 1.007825u and m_n = 1.008665u.

Answer 1

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,

N_A = 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 × m_H + 34 × m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 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 = Δmc²

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

= 1.581 × 10²⁵ MeV

But 1 MeV = 1.6 × 10⁻¹³ J

E_b = 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.

Answer 2

Number of atoms in 3 gram of Cu coin = `((6.023 xx 10^23 xx 3))/63 = 2.86 xx 10^22`

Each atom has 29 Protons & 34 Neutrons.

Thus Mass defect Δm = `29 xx 1.00783 + 34 xx 1.00867 - 62.92960` u = 0.59225 u

Nuclear energy required for one atom = `0.59225 xx 931.5` MeV

Nuclear energy required for 3 gram of Cu = `0.59225 xx 931.5 xx 2.86 xx 10^22 "MeV" = 1.58 xx 10^25` MeV