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