Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The binding energy, $ B $, of an electron in a hydrogen atom is given as $ 13.6 \, \text{eV} $. This energy is what keeps the electron bound to the proton, effectively reducing the total mass of the hydrogen atom compared to the sum of a free proton and electron.
Now we need to calculate the mass decrease by using Einstein’s relation $ B=\Delta m C^2$, so
$$
\Delta m =\dfrac{B}{c^2}
$$
Plug the known: where $ c^2 = 931.49 \, \text{MeV}/\text{u} $,
$$ \Delta m = \frac{13.6 \, \text{eV}}{931.49 \, \text{MeV } } \times 1 \, \text{u}$$
$$
\Delta m = \frac{13.6 }{931.49\times 10^6 } \times 1 \, \text{u}= \color{red}{\bf 1.46 \times 10^{-8} }\,\rm \text{u}
$$
The mass of a hydrogen atom is approximately $ 1.007825 \, \text{u} $, so
$$
\frac{\Delta m}{m_{\rm H}} = \frac{1.46 \times 10^{-8} \, \text{u}}{1.007825 \, \text{u}} \times 100\%=\color{red}{\bf 1.45 \times 10^{-6}}\%
$$
$$\color{blue}{\bf [b]}$$
In this case, we’re looking at the formation of a helium nucleus from two protons and two neutrons. The mass defect, or the amount by which the mass decreases, can be calculated using nuclear mass values.
The formula for mass change $ \Delta m $ in forming helium is given by
$$
\Delta m = 2m_p +2 m_e+ 2m_n - (m_{\text{He}}+2e)
$$
where $ m_p $ is the proton mass, $ m_e $ is the electron mass, $ m_n $ is the neutron mass, and $ m_{\text{He}} $ is the mass of the helium nucleus.
$$
\Delta m = 2m_{\rm H} + 2m_n - (m_{\text{He}}+2e)
$$
where $ (m_p+m_e=m_{\rm H}) $ which is the hydrogen atom mass.
Plug the known;
$$
\Delta m = 2(1.007825 \, \text{u}) + 2(1.008665 \, \text{u}) -( 4.002602 \, \text{u}+0.0005486\,\text{u})$$
$$
\Delta m =\color{red}{\bf 0.0299}\, \rm {u}
$$
The mass of a helium nucleus is about $ 4.0026 \, \text{u} $, so the percentage of mass loss is:
$$
\frac{\Delta m}{m_{\text{He}}} = \frac{0.0299\, \text{u}}{4.0026 \, \text{u}} \times 100\%= \color{red}{\bf 0.75}\%
$$
$$\color{blue}{\bf [c]}$$
Why Mass Loss is Significant in Nuclear Reactions but Not in Chemical Reactions
In chemical reactions, the binding energies involved (like the 13.6 eV in a hydrogen atom) are typically very small compared to the masses of the particles. This results in an extremely tiny mass change, as calculated in Part (a), which is often negligible and challenging to detect experimentally.
In nuclear reactions, however, the binding energies are much larger (often in the range of MeV rather than eV). This substantial binding energy leads to a more noticeable mass defect, as seen in the helium nucleus calculation in Part (b), where the mass loss is nearly 1% of the nucleus's total mass. This large change is both measurable and crucial to the study of nuclear reactions, as it directly correlates with the energy released or absorbed during these reactions.
