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).

(2) Hint : There is reduction in the mass as some mass is lost in the form of energy .

Step 1 : Find the value of M .

Given,

F = \frac{GMm}{r^{2}}

M = effective mass of hydrogen atom

= mass of electron + mass of proton - \frac{B^{2}}{C}

where B is BE of hydrogen atom = 13.6 eV

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