Ncert Exercises & Solutions

The gravitational force between \(\text H\text-\)atom and another particle of mass \(m\) will be given by Newton's law \(F=\dfrac{GMm}{r^2},\) where \(r\) is in \(\text{km}\) and;

1.	\(M = m_{\text{proton}}+ m_{\text{electron}}.\)
2.	\(M = m_{\text{proton}}+ m_{\text{electron}}-\frac{B}{c^2}\left(B= 13.6~\text{eV}\right)\).
3.	\(M\) is not related to the mass of the hydrogen atom.
4.	\(M = m_{\text{proton}}+ m_{\text{electron}}-\frac{\|V\|}{c^2}(\|V\|=\) magnitude of the potential energy of electron in the \(\text H\text-\)atom).

Hint: The mass defect is given by; \(\Delta M = \frac{B}{C^2}\)

Step: Find the net effective mass.
Given,
\(F=\frac{GMm}{r^2} \)
The net effective mass is given by;
Total mass = mass of proton + mass of electron -mass defect
The mass defect is given by ;
\(\Rightarrow \Delta M = \frac{B}{C^2}\)
where binding energy \(B=13.6~\text{eV} ,\) \(c\) is the speed of light.
Substituting this value, we get;
\(\Rightarrow M=m_{proton} + m_{electron} -\Delta m\)
\(\Rightarrow M = m_{\text{proton}}+ m_{\text{electron}}-\frac{B}{c^2}\)
Hence, option (2) is the correct answer.

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