For an electron revolving in a hydrogen atom, the electrostatic force provides the necessary centripetal force: $$ \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} $$ Rearranging for the radius $r$: $$ r = \frac{1}{4\pi\epsilon_0} \frac{e^2}{mv^2} $$
Given: $v = \sqrt{25.6} \times 10^5 \, ms^{-1} \implies v^2 = 25.6 \times 10^{10} \, m^2s^{-2}$ $m = 9 \times 10^{-31} \, kg$ $e = 1.6 \times 10^{-19} \, C$ $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, Nm^2C^{-2}$
Substitute the values into the formula: $$ r = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(9 \times 10^{-31}) \times (25.6 \times 10^{10})} $$ $$ r = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-31} \times 25.6 \times 10^{10}} $$ $$ r = \frac{2.56 \times 10^{-29}}{25.6 \times 10^{-21}} $$ $$ r = 0.1 \times 10^{-8} \, m = 1 \times 10^{-9} \, m $$
Comparing $r = 1 \times 10^{-9} \, m$ with $x \times 10^{-9} \, m$, we find $x = 1$.
Final Answer: 1
AI generated content. Review strictly for academic accuracy.