Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5\ \text{m s}^{-1}$. The radius of the orbit is $x \times 10^{-9}$ m. The value of $x$ is :
[Take the mass of electron to be $9 \times 10^{-31}$ kg, charge of electron $= -1.6 \times 10^{-19}$ C and $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2\text{C}^{-2}$]
Answer: (D) $1$
The Coulomb force provides the centripetal force:
$$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2} \;\Rightarrow\; r = \frac{1}{4\pi\epsilon_0}\frac{e^2}{mv^2}$$
$$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times 25.6 \times 10^{10}} = \frac{9 \times 2.56 \times 10^{-29}}{9 \times 2.56 \times 10^{-20}} = 10^{-9}\ \text{m}$$
So $x = 1$.
Solution by Sreeraj P, M.Sc Physics