Ncert Exercises & Solutions

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

Hint: Mass defect \(\Delta M = \frac{B}{C^2}\)

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

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