Physics - Modern Physics Question with Solution | TestHub

The electrostatic force between the nucleus and the electron provides the necessary centripetal force for the electron to revolve in its orbit.

$\frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}$

Rearranging for the radius $r$:

$r=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{m{v}^2}$

Substituting the given values:
$\frac{1}{4\pi {ϵ}_0}=9\times 10^9$ N m${}^2$ C${}^{-2}$
$e=1.6\times 10^{-19}$ C
$m=9\times 10^{-31}$ kg
$v=\sqrt{25.6}\times 10^5$ m s${}^{-1}$

$r=\frac{9\times 10^9\times (1.6\times 10^{-19}{)}^2}{9\times 10^{-31}\times (\sqrt{25.6}\times 10^5{)}^2}$

$r=\frac{9\times 10^9\times 2.56\times 10^{-38}}{9\times 10^{-31}\times 25.6\times 10^{10}}$

$r=\frac{2.56\times 10^{-29}}{25.6\times 10^{-21}}$

$r=0.1\times 10^{-8}$ m

$r=1\times 10^{-9}$ m

Comparing this with $r=x\times 10^{-9}$ m, we get:

$x=1$

Answer: $1$