Question

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

Answer 1

HintThe 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)}}$