physics

Question

Find the binding energy in MeV for radium (atomic mass = 226.0254 u). The mass of a proton

is 1.007276u and the mass of a neutron is 1.008665u. Note: you do not need to include the mass

defect of electrons.

Answer 1

(1)

The nuclear mass of the Radium is 226.0254 u

In a radium, there are 88 protons/Electrons and 138 neutrons.

The mass of each proton is 1.007276 u

The mass of each electron is 1.008665 u

Speed of the light is 3 * 10⁸ m/sec.

 

Mass defect:

The difference between the mass of the atom and the sum of the masses of individual proton, electron and neutron is called  the  mass  defect  (Dm).

Dm = (Number of Proton * mass of each proton) + (Number of Neutron * mass of each Neutron) - Mass of the atom

Dm = (88*1.007276) + (138 * 1.008665) - (226.0254)

Dm = 88.640 + 139.19577 - 226.0254

Dm =  1.81066 amu

We know that 1 amu = 1.66053892 × 10^-27 kilograms

Dm =  1.81066 * 1.66053892 × 10^-27 kilograms

Dm =  3.00667 * 10^-27 kg

Dm =  3.007 * 10^-27 kg

The mass defect of the Radium is  3.007 * 10^-27 kg.

 

Binding energy = mC²

Binding energy = 3.007 * 10^-27 * (3 * 10⁸)²

E = 2.706 * 10^-10 J

We know that 1Mev = 1.6 * 10^-13 J ⇒ 1J = 1/(1.6 * 10^-13)

E = 2.706 * 10^-10 * (1/(1.6 * 10^-13)) = 1691.25

Therefore binding energy of Radium is 1691.25 Mev.