Question: The masses of neutrons and protons are 1.0087 and 1.0073 amu respectively. If the neutrons and proto...

Question

The masses of neutrons and protons are 1.0087 and 1.0073 amu respectively. If the neutrons and protons combine to form helium nucleus of mass 4.0015 amu the binding energy of the helium nucleus will be
(A) 28.4 MeV
(B) 20.8 MeV
(C) 27.3 MeV
(D) 14.2 MeV

Answer 1

Solution

Hint
Binding energy is the energy that would be liberated by combining the individual protons and neutrons into a single nucleus. mathematically it is calculated as, $B.E. = \Delta m{c^2}$ , ∆m is the mass defect and c is the velocity of light.

Complete step by step answer
We can find out the binding energy by using the formula $B.E. = \Delta m{c^2}$
Where, ∆m is the mass defect
C is the velocity of light
Here, it is given that
Mass of neutron = 1.0087amu
Mass of proton = 1.0073 amu
As, in helium nucleus there are 2 protons and 2 neutrons
So theoretical mass is = $\left( {2 \times 1.0087 + 2 \times 1.0073} \right)$
= $4.032a.m.u$
Now, mass of nucleus is = 4.0015a.m.u
So mass defect is given by
$\Rightarrow \Delta m = \left( {4.032 - 4.0015} \right)a.m.u$
$\Rightarrow \Delta m = 0.0305a.m.u$
Now we know that the binding energy is $B.E. = \Delta m{c^2}$
$B.E. = \left( {0.0305a.m.u} \right){c^2}$
And $1a.m.u = 931{{MeV} \mathord{\left/ {\vphantom {{MeV} {{c^2}}}} \right.} {{c^2}}}$
Therefore, binding energy is $B.E. = \left( {0.0305} \right) \times 931MeV$
$\Rightarrow B.E. = 28.4MeV$ .
Hence, (A) option is correct.

Additional Information
This is true that the mass of the nucleus is little less than the mass of individual neutrons and protons. This missing mass is known as mass defect and represents the binding energy of the nucleus.

Note
The binding energy can be calculated by adding the masses of individual protons, neutrons and electrons and subtracting the mass of the atom and then convert the difference into energy. We must keep in mind that the value of 1a.m.u in units ${{MeV} \mathord{\left/ {\vphantom {{MeV} {{c^2}}}} \right.} {{c^2}}}$ is $931{{MeV} \mathord{\left/ {\vphantom {{MeV} {{c^2}}}} \right.} {{c^2}}}$ so by putting this value we can answer easily.