Question

The masses of $\alpha -$ particle, proton and neutron are $4.0015amu$ and $1.00867amu$ respectively. Binding energy per nucleon of $\alpha -$ particle is:
(A) $283J$
(B) $931.5MeV$
(C) $28.3MeV$
(D) $7.08MeV$

Answer 1

Hint: Let us remember that a nucleus contains particles like protons and neutrons, but for such particles there are also some repulsive forces that live amongst the protons' positive charges. Binding energy is used to keep the nucleons as a combined unit within the nucleus.

Complete answer:
An atom's nucleus can be described to be the core region of any atom that contains the bulk of its mass.
The term nucleon refers to the category of subatomic particles that reside in the nucleus, including protons and neutrons. In this case, the alpha particle is actually composed of two protons and two neutrons.
The 'nuclear binding energy' is said to be the least quantity of strength or energy that is needed to disassemble/break down the nucleus of the given atom into the subatomic particles that make up the nucleus (neutrons and protons, to specify).
Let us write down the given values:
Mass of alpha particle (a) $=4.0015amu$
Mass of proton (p) $=1.00728amu$
Mass of neutron (n) $=1.00867amu$
Also note that the energy corresponding to each atomic mass unit (Energy) $=931.478MeV$
Now we need to know certain formulas in order to solve this question:
Binding energy is the product of the Mass defect and energy of each atomic mass unit.
 $\Rightarrow B.E.=M.D\times Energy$ ; where B.E. is binding energy, M.D. is mass defect and Energy is the energy corresponding to a single atomic mass unit.
So to find the binding energy, we require the mass defect.
Mass defect is the difference between the original mass of an alpha particle and the mass of an alpha particle that is expected.
 $\Rightarrow M.D.=Expectedmass-Actualmass$
We can proceed to the calculations:
To calculate expected mass of the alpha particle we know its constituent particles.
 $Expectedmass=2\times p+2\times n$ ; here ‘ $p$ ’ is the mass of a proton and ‘ $n$ ’ is the mass of a neutron.
 $\Rightarrow Expectedmass=[2\times 1.00728+2\times 1.00867]amu$
 $\Rightarrow Expectedmass=4.0319amu$
We already know the actual mass which is the given mass of the alpha particle:
  $\Rightarrow Actualmass=4.0015amu$
Now we can compute the mass defect as:
 $\Rightarrow M.D.=4.0319amu-4.0015amu$
 $\Rightarrow M.D.=0.0304amu$
Next we can calculate the binding energy by substituting the values of mass defect and energy of a single atomic mass unit;
 $\Rightarrow B.E.=0.0304amu\times 931.478MeV$
 $\Rightarrow B.E.=28.32MeV$
To find binding energy per nucleon we need the nucleon number;
Nucleon number for the alpha particle is the sum of the number of neutrons and protons in the alpha particle.
 $\Rightarrow Nucleonnumber=2+2=4$
Therefore binding energy per nucleon will be:
 $\Rightarrow {\displaystyle \frac{B.E.}{Nucleonnumber}}={\displaystyle \frac{28.32MeV}{4}}$
 $\Rightarrow {\displaystyle \frac{B.E.}{Nucleonnumber}}=7.08MeV$
For the alpha particle, the binding energy per nucleon is calculated to be $7.08MeV$ .
So from the above calculation it is evident that the options (A), (B) and (C) containing values 283J, 931.5MeV and 28.3 MeV respectively can be rejected since they do not contain the calculated value. But option (D) contains the right value since it has the value .
Therefore the correct answer is clearly option (D) .

Note:
As an atom radioactively decays, it releases certain particles such as rays of alpha, beta, gamma. Radioactivity happens mostly as an unstable atom attempts to achieve equilibrium. As a result, when they become unstable, they decay by releasing a particle that transforms the nucleus into its alternative form or moves into an energy state that is lower. This sequence of decays continues until the nucleus reaches a stable state.