Question

Calculate mass defect in the following reaction:
$_{1}{{H}^{2}}{{+}_{1}}{{H}^{3}}{{\xrightarrow{{}}}_{2}}H{{e}^{4}}{{+}_{0}}{{n}^{1}}$
(Given: Mass $_{1}{{H}^{2}}$ = 2.014 amu, $_{1}{{H}^{3}}$ = 3.016 amu, $_{2}H{{e}^{4}}$ =4.004 amu, $_{0}{{n}^{1}}$ =1.008 amu)
(A) 0.018 amu
(B) 0.18 amu
(C) 0.0018 amu
(D) 1.8 amu
(E) 18 amu

Answer 1

Hint: Firstly, we should know about the mass defect. A mass defect is the difference between the actual mass of an atom and the total mass of protons and neutrons forming a nucleus. It is represented as ∆m.

Formula Used: The total mass of the reactant $\left( {{M}_{r}} \right)$ and the total mass of products $({{M}_{p}})$ are calculated. Their difference is then calculated, which is equal to the mass defect. The mass defect is also known as excess mass.
Mass defect = ${{M}_{r}}-{{M}_{p}}$

Complete Step by Step Solution:
The mass defect is given by
$\Delta m=[Z({{m}_{p}}+{{m}_{e}})+(A-Z){{m}_{n}}]-M$
Where ${{m}_{p}}$is the mass of a proton (1.007277 amu)
             ${{m}_{e}}$is the mass of electron (0.000548597 amu)
             ${{m}_{n}}$is the mass of neutron (1.008665 amu)
             $M$is the atom’s atomic mass
             $Z$is the atom’s atomic number
The masses of the reactants and products are $_{1}{{H}^{2}}$= 2.014 amu, $_{1}{{H}^{3}}$ = 3.016 amu, $_{2}H{{e}^{4}}$=4.004 amu, and $_{0}{{n}^{1}}$=1.008 amu, respectively.
Total mass of reactants (${{M}_{r}}$) = mass of $_{1}{{H}^{2}}$ + mass of$_{1}{{H}^{3}}$= 2.014+3.016 = 5.030 amu
Total mass of products (${{M}_{p}}$) = mass of $_{2}H{{e}^{4}}$ + mass of $_{0}{{n}^{1}}$= 4.004 + 1.008 = 5.012 amu
Mass defect = ${{M}_{r}}-{{M}_{p}}$
                      = 5.030 – 5.012
                      = 0.018 amu
Correct Option: (A) 0.018 amu.

Note: It must be remembered that mass defect means that the masses of the components of an atom are different from the actual mass of the atom. The reason behind it is that some of the mass is released as energy when neutrons and protons bind in the atomic nucleus.