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Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6}\times10^5$ ms$^{-1}$. The radius of the orbit is $x\times10^{-9}$ m. The value of x is: [Take the mass of electron to be $9\times10^{-31}$ kg, charge of electron = $-1.6\times10^{-19}$ C and $\frac{1}{4\pi\varepsilon_0}=9\times10^9$ Nm$^2$C$^{-2}$]
Detailed Solution
$\frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \Rightarrow r=\frac{9\times10^9\times(1.6\times10^{-19})^2}{9\times10^{-31}\times25.6\times10^{10}}=\frac{256\times10^{-40}}{256\times10^4\times10^5}\times\frac{10^9}{10^{-31}}=1\times10^{-9}$ m. So x = 1.