Question

The mass defect of ${}_2^4He\,$is $0.03\,u$the binding energy per nucleon of Helium (in MeV) is:

Answer 1

Hint The given problem is related with the following point:

1. Nuclear binding energy
2. Mass defect
3. Binding energy per nucleon
   The knowledge of these points will help to solve the problem.

Complete step-by-step solution :First we must have the knowledge of binding energy. It is the energy required to break the nucleus.
The mass defect defined as: The mass of each element atom is less than the sum of masses of its constituent particles
In the given problem the atom is $\left[ {{}_2^4He} \right]$ for the calculation of mass defect, find the difference between the mass of the atom and the total mass of its constituents.
Since the mass of the constituents is:
[mass of two H-atoms $+$ mass of two neutrons]
Again the meaning of binding energy per nucleon is that it is energy to remove a nucleon from elements.
Now binding energy per nucleon$= \dfrac{{\vartriangle m.{c^2}}}{A}$
Here mass defect $= \vartriangle m = 0.03\,u$
$1u$ is equivalent to $\left( {931.5\,\,MeV} \right)$
Binding energy is $\vartriangle m.{c^2}$
Or binding energy $=$ (mass defect $\times 931.5$) MeV $= \left( {0.03 \times 931.5} \right)$ MeV
Therefore binding energy per nucleon
$= \dfrac{{\;0.03 \times 931.5}}{4} = 6.986$ MeV

Note: It is necessary to have the knowledge of following terms

1. Mass defect: $1\,amu/1u = 1.656 \times {10^{ - 27}}\,kg$
   Or $\vartriangle E = \vartriangle M \times {c^2} = \left( {1.656 \times {{10}^{ - 27}}} \right){\left( {3 \times {{10}^8}} \right)^2}\,J = 931.5\,MeV$
2. Binding energy per nucleon $= \dfrac{{\left( {mass\,defect} \right)\left( {{c^2}} \right)}}{{nucleon\,number\,\left( A \right)}}$