Question: Q. 1. The gravitational force between a $\mathrm{H}$-atom and another particle of mass $m$ will be given by Newton’s law :
$F=G \frac{M . m}{r^{2}}$, where $r$ is in $\mathrm{km}$ and
(a) $M=m_{\text {proton }}+m_{\text {electron }}$
(b) $\quad M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^{2}} \quad(B=13.6 \mathrm{eV})$.
(c) $M$ is not related to the mass of the hydrogen atom.
(d) $\quad M=m_{\text {proton }}+m_{\text {electron }}-\frac{|V|}{c^{2}}(|V|=\quad$ magnitude of the potential energy of electron in the $\mathrm{H}$-atom).
[NCERT Exemplar]
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Solution:
Ans. Correct option : (b)
Explanation: During formation of $\mathrm{H}$-atom some mass of nucleons convert into energy by the equation
$$ E=m c^{2} $$
This energy is used to bind the nucleons along with nucleus. So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons.
Actual mass of $\mathrm{H}$ atom
$$ =M_{p}+M_{e}-\frac{B . E .}{c^{2}} \quad\left(\frac{B}{c^{2}}=\text { Binding energy }\right) $$
So, the binding energy of $\mathrm{H}$ atoms is $13.6 \mathrm{eV}$ per atom.
