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 ( text where k=(1/4 π ε0))

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 ( text where k=(1/4 π ε0)) (A)  k ( e2 / r3 ) r (B) - k ( e2 / r3 ) r (C)  k ( e2 / r ) r (D)  - k ( e2 / r ) r

Tardigrade Admin · Accepted Answer

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 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} r

Putting,

\frac{1}{4 π ε _{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 a l so 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

Q. 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
$(where k = \frac{1}{4 π ε _{0}})$

$k \frac{e ^{2}}{r ^{3}} r$

$- k \frac{e ^{2}}{r ^{3}} r$

$k \frac{e ^{2}}{r} r$

$- k \frac{e ^{2}}{r} r$

Q. 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 ( where k=4πε0​1​)

kr3e2​r

−kr3e2​r

kre2​r

−kre2​r

Q. 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
$(where k = \frac{1}{4 π ε _{0}})$

$k \frac{e ^{2}}{r ^{3}} r$

$- k \frac{e ^{2}}{r ^{3}} r$

$k \frac{e ^{2}}{r} r$

$- k \frac{e ^{2}}{r} r$