Question

What is meant by binding energy? If the masses of proton, neutron and alpha $\left( \alpha \right)$ particles are $1.00728\,amu,\,1.00867\,amu$ and $4.00150\,amu$ respectively, then find the binding energy per nucleon of alpha particle. $\left[ {1\,amu = 931\,MeV} \right]$.

Answer 1

Hint: The difference of mass of product with reactants is known as mass decay. If mass decay into a nuclear reaction is $\Delta m$ then the binding energy is given by B.E. $ = \Delta m \times 931\,MeV$and the binding energy per nucleon means the ratio of binding energy with total number of nucleons present into the nucleus of an element.

Complete step by step answer:
Binding Energy-The amount of energy is required to separate a particle from a system of particles is known as binding energy.
The binding energy is basically applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.
Part 2: Given that,
mass of proton ${m_p} = 1.00728\,amu$
mass of neutron ${m_n} = 1.00867\,amu$
mass of $\alpha - $particle ${m_\alpha } = 4.00150\,amu$
The $\alpha - $particle is basically the nucleus of the Helium atom. The notation of $\alpha - $particle is $_2H{e^4}$, which has two protons and two neutrons,
When $\alpha - $particle breaks, the reaction is following: $_2H{e^4} \to 2p + 2n$ where “p” presents protons and “h” represents neutrons.
Now, the difference in the mass of product and reactant is known as decay of mass, say$\Delta m$
So $\Delta m = $Total mass of product $ - $Total mass of reactants
$\Delta m = \left[ {2 \times {m_p} + 2 \times {m_n}} \right] - {m_\alpha }$
$ = \left[ {2 \times 1.00728 + 2 \times 1.00867} \right]$
$ - \left[ {4.00150} \right]$
$\Delta m = 0.0304\,amu$
Now, using the formula of binding energy,
$BE = \Delta m \times 931\,MeV$
$BE = 0.0304 \times 931\,MeV$
$BE = 28.3024\,MeV$
We have to find binding energy per nucleon, the total nucleons present in $\alpha - $particles are, (two protons $ + $two neutron) that means, Total number of nucleons$ = 2p + 2n = 4$.
So the binding energy per nucleon of the $\alpha - $particle is,
$\dfrac{{Binding\,\,Energy}}{{Tota\operatorname{l} \,\,number\,\,of\,\,nucleons}} = \dfrac{{28.3024}}{4}$
$ = 7.0756\,MeV$

Note:
To calculate the total number of nucleons in an element, let us consider an element \[_z{B^A}\]where B is symbol of the element, Z is atomic number of the element B and A is mass number of the element B. Now, the total number of nucleons is sum of, (protons $ + $neutrons) present in the nucleus of the element B. Here, the number of protons $ = $mass number $ = $Z and number of neutrons $ = $A $ - $Z. So the number of nucleons present into the nucleus of element B is $ = $Z $ + $(A $ - $Z).