Question

An alpha particle $({}_{4}^{2}He)$ has a mass of 4.00300 amu. A proton has a mass of 1.00783 amu and a neutron has a mass of 1.00867 amu respectively. The binding energy of alpha particle estimated from these data is the closest to
A. 27.9MeV
B. 22.3 MeV
C. 35.0 MeV
D. 20.4 MeV

Answer 1

To solve this question apply the concept of mass defect and binding energy. We will find the mass defect using its formula. Afterward, we will use the formula for binding energy and hence obtain the answer.

Formula used:
Mass defect
$\mathbf{\Delta }M=2{{m}_{p}}+2{{m}_{n}}-{{m}_{he}}$

Complete step by step answer:
An alpha particle $({}_{4}^{2}He)$ has 2 protons and two neutrons.
We know the formula of the mass defect.
Mass defect
$\mathbf{\Delta }M=2{{m}_{p}}+2{{m}_{n}}-{{m}_{he}}$
Putting values respectively.
$\therefore ~~~\Delta M=2(1.00783)+2(1.00867)-4.00300~~=0.03000~amu$,
We also know that

Binding energy of alpha particle
$B.E=\Delta M{{c}^{2}}$

& ~(\because 1~amu~~=931.5MeV~) \\\ & ~\therefore ~~\text{ }B.E~=0.03000\times 931.5=27.9~MeV \\\ \end{aligned}$$ The binding energy of alpha particles estimated from these data is the closest to 27.9 MeV. **So the correct option is (A).** **Additional Information:** Let us take a glimpse of a particle. A particle is a quick-moving packet containing two protons and two neutrons or we will say a helium nucleus. Alpha particles carry a charge of +2 and strongly interact with matter. Alpha particles can travel only a couple of inches through the air and maybe easily stopped with a sheet of paper. A particle is produced by a process decay. **Note:** We know that separation energy is the amount of energy required to separate a particle from a system of particles or we will say that energy required to disperse all the particles of the system is separation energy. While calculating the mass deficiency it's important to use the complete accuracy of mass measurements. Because the difference in mass is little compared with the mass of the atom. Rounding off the masses of atoms and particles to 3 or four significant digits before the calculation will result in zero mass deficiency.