A) \[k\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r}\]
B) \[-k\,\frac{{{e}^{2}}}{{{r}^{3}}}\vec{r}\]
C) \[k\frac{{{e}^{2}}}{{{r}^{2}}}\hat{r}\]
D) \[-k\frac{{{e}^{2}}}{{{r}^{3}}}\hat{r}\]
Correct Answer: B
Solution :
Let charges on an electron and hydrogen nucleus are \[{{q}_{1}}\] and \[{{q}_{2}}\]. The Coulomb's force between them at a distance r is, \[\vec{F}=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\hat{r}\] Putting \[\frac{1}{4\pi {{\varepsilon }_{0}}}=k\,(given)\] \[\vec{F}=-k\,\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}}\,\hat{r}\] Since, the nucleus of hydrogen atom has one proton, so charge on nucleus is e i.e., \[{{q}_{2}}=e\] also \[{{q}_{1}}=e\] for electron \[So,\,\vec{F}=-k\,\frac{e\,.\,e}{{{r}^{2}}}\hat{r}=-k\frac{{{e}^{2}}}{{{r}^{2}}}\,\hat{r}\] \[but\,\,\hat{r}=\frac{{\vec{r}}}{|\vec{r}|}=\frac{{\vec{r}}}{r}\] Hence, \[\vec{F},\,=-k\frac{{{e}^{2}}}{{{r}^{2}}}.\frac{{\vec{r}}}{r}=-k\frac{{{e}^{2}}}{{{r}^{3}}}.\vec{r}\] Note: Negative sign in the expression for Coulomb?s force shows that force between electron and hydrogen nucleus is of attraction. (where \[k=\frac{1}{4\pi {{\varepsilon }_{0}}}\])
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