An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force \overrightarrow{F} between the two is - (where {\K}=\frac{1}{4 \pi \epsilon_0} )

Correct Answer is : −Ke2r3→r

{\K} { {\e}^2}/{ {\r}^2} \hat{ {\r}

- {\K} { {\e}^2}/{ {\r}^3} \hat{ {\r}

K \frac{e^2}{r^3} \overrightarrow{r}

-K \frac{e^2}{r^3} \overrightarrow{r}

AIPMT 2003 Physics and Chemistry Paper

Question : 25 of 100

Marks: +1, -0

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force

→

between the two is -

(where K=‌

4πε0

)

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