Find the energy liberated in the reaction

²²³Ra → ²⁰⁹Pb + ¹⁴C.

The atomic masses needed are as follows.

²²³Ra ²⁰⁹Pb ¹⁴C

223.018 u 208.981 u 14.003 u

(a) Calculate the energy released if ²³⁸U emits an α-particle.

(b) Calculate the energy to be supplied to ²³⁸U if two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is

ΔE = (M_{z–1, N} +M_H –M_{Z, N}) c²

Where M_{Z, N} = mass of an atom with Z protons and N neutrons in the nucleus and M_H = mass of a hydrogen atom. This energy is known as proton-separation energy.

Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons in terms of the masses M_{Z, N}, M_{Z, N–1} and the mass of the neutron.