Question

Find the binding energy of an $\alpha$- particle from the following data.
Mass of the helium nucleus$=4.001265a.m.u$
Mass of the proton $=1.007277a.m.u$
Mass of the neutron $=1.00866a.m.u$
Take $1a.m.u=931.4813MeV$

Answer 1

We need to understand the relation between the mass of the helium atom, the mass of the neutron and the proton and the binding energy of the alpha particle to solve the problem. We should consider the helium atom and the alpha particle to be the same.

Complete answer:
We know that the alpha particle is basically the helium nucleus with an atomic mass of 4 u. It consists of two protons and two neutrons.
We know that the sum total of the masses of the two protons and the two neutrons should be the mass of the nucleus. But from our experiences with nuclear studies, we know that there is a loss of some mass when we weigh a nucleus as a single entity. i.e., the mass of the nucleus will be lesser than the sum total of the masses of individual protons and neutrons.
This anomaly in the mass is explained by the nuclear binding energy involved in each nucleus. The nucleus being a highly positive entity requires some external energy to hold the nucleons in position. This energy is spent on the mass defect that we found from experiments.
We are given the mass of the helium nucleus, the proton and the neutron. We can find the binding energy of the alpha-particle very easily using the following formula as –

& BE=\Delta M{{c}^{2}} \\\ & \Rightarrow \Delta M=\\{(2{{M}_{p}}+2{{M}_{n}})-{{M}_{\alpha }}\\} \\\ & \Rightarrow \Delta M=\\{(2\times 1.007277a.m.u)-(2\times 1.00866a.m.u)-4.001265a.m.u\\} \\\ & \Rightarrow \Delta M=\\{2.014554-2.01732-4.001265\\} \\\ & \Rightarrow BE=\\{0.030609\\}(931.4813MeV) \\\ & \therefore BE=28.511MeV \\\ \end{aligned}$$ The Binding energy of the Alpha particle is 28.511MeV. This is the required solution. **Note:** The binding energy of a nucleus is the measure of the strength of the nucleus. It is a positive quantity because we need to provide the energy to split up the nucleus into its respective nucleons as we do even in the nuclear reactions to start the chain reaction.