Question

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is
$\left(\text { where } k=\frac{1}{4 \pi \varepsilon_0}\right)$

• A. $ k \frac{ e^2 }{ r^3 } r$
• B. $- k \frac{ e^2 }{ r^3 } r$
• C. $ k \frac{ e^2 }{ r } r$
• D. $ - k \frac{ e^2 }{ r } r$

Answer 1

Answer: (B)

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,
F = - $\frac{1}{ 4 \pi \varepsilon_0} \frac{ q_1 q_2 }{ r^2} r$
Putting, $ \frac{1}{ 4 \pi \varepsilon_0} = k$ (given)
F = - k $ \frac{ q_1 q_2 }{ r^2} 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, $ F = - \frac{ e.e}{ r^2 } r = - k \frac{e^2 }{ r^2 } r$
but r = $ \frac{ r}{ | r |} = \frac{ r}{ r } $
Here, $F = - k \frac{e^2 }{ r^2 }. \frac{ r}{ r } = - \frac{e^2 }{ r^3} . r$