Question

A helium atom, a hydrogen atom and a neutron have masses of 4.003u, 1.008u and 1.009u (unified mass units), respectively. Assuming that hydrogen atoms and neutrons can fuse to form helium, what is the binding energy of the helium nucleus?
A. 2.01u
B. 3.031u
C. 1.017u
D. 0.031u

Answer 1

Knowing the formula of the binding energy of an atom ${}^{A}{{X}_{Z}}$, where the mass of the atoms is M is; BE=\left\\{ \left[ Zm\left( {}^{1}H \right)+N{{m}_{n}} \right]-m\left( {}^{A}X \right) \right\\}{{c}^{2}}. In this formula, $m\left( {}^{1}H \right)$ is the mass of Hydrogen atom and $\left( {{m}_{n}} \right)$ refers to the mass of neutron and $m\left( {}^{A}X \right)$ is the mass of the atom X.

Step by step solution:
Let’s start by finding out what is Binding Energy of a nucleus. The binding energy of a nucleus is defined as the energy required to completely disassemble the nucleus into its constituent protons and neutrons. The bound system of the nucleus has smaller mass than its constituents, hence this difference in the mass is known as mass defect:$\Delta m$. This mass defect will be converted into energy using Einstein's relation: $BE=\left( \Delta m \right){{c}^{2}}$.
For an atom${}^{A}{{X}_{Z}}$, whose nucleus contains Z protons and N neutrons. Hence, the mass defect becomes: $\Delta m=\left( Z{{m}_{p}}-N{{m}_{n}} \right)-{{m}_{total}}$. Thus, the binding energy becomes; $BE=\left( \Delta m \right){{c}^{2}}\Rightarrow BE=\left[ \left( Z{{m}_{p}}-N{{m}_{n}} \right)-{{m}_{total}} \right]{{c}^{2}}.$ Adding and subtracting Z times the mass of electrons to the mass defect, we get; $\Delta m=\left( Z{{m}_{p}}+Z{{m}_{e}}-N{{m}_{n}} \right)-\left( {{m}_{total}}+{{m}_{e}} \right)\Rightarrow \Delta m=\left[ Zm({}^{1}H)-N{{m}_{n}} \right]-m\left( {}^{A}X \right)$.
Therefore, the updated formula of binding energy becomes; BE=\left\\{ \left[ Zm\left( {}^{1}H \right)+N{{m}_{n}} \right]-m\left( {}^{A}X \right) \right\\}{{c}^{2}}.
In the current problem, we have a mass of Helium atoms at 4.003u. That is; $m\left( {}^{4}He \right)=4.003u$. The mass of the Hydrogen atom is 1.008u. That is; $m\left( {}^{1}H \right)=1.008u$ and the mass of the neutron is 1.009u. That is; ${{m}_{n}}=1.009u$.
We know that the Helium nucleus consists of 2 protons, 2 electrons and 2 neutrons. Hence Z=2 and N=2 for Helium.
Hence, the Binding energy of Helium will be; BE\left( {}^{4}He \right)=\left\\{ \left[ 2m\left( {}^{1}H \right)+2{{m}_{n}} \right]-m\left( {}^{4}He \right) \right\\}{{c}^{2}}.
Substituting in the above values of mass of Helium, mass of Hydrogen and mass of neutron in the above equation, we get; \begin{aligned} & BE\left( {}^{4}He \right)=\left\\{ \left[ 2\left( 1.008u \right)+2(1.009u) \right]-m\left( 4.003u \right) \right\\}{{c}^{2}} \\\ & \Rightarrow BE\left( {}^{4}He \right)=\left\\{ \left[ 2\left( 1.008u \right)+2(1.009u) \right]-\left( 4.003u \right) \right\\}{{c}^{2}}\Rightarrow BE\left( {}^{4}He \right)=\left( 0.032u \right){{c}^{2}} \\\ \end{aligned}
Hence, the binding energy of the Helium nucleus is given by Option D.

Note: The process of disintegrating a heavy nucleus into its lighter daughter nuclei is known as the process of nuclear fission. During this process of disintegration, energy is released due to the difference in the binding energy of the heavy and the lighter daughter nuclei.
The above process is an example of nuclear fusion reaction, where 2 protons and 2 neutrons combine to form a helium nucleus. During this process, energy released is given by the binding energy.