In the Bohr model of the hydrogen atom,the ce | vedclass

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(C) In the Bohr model,the centripetal force required for the circular motion of the electron is provided by the electrostatic Coulomb force of attract

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$\frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$

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(C) In the Bohr model,the centripetal force required for the circular motion of the electron is provided by the electrostatic Coulomb force of attraction between the nucleus (proton) and the electron.The condition for equilibrium is:$	ext{Centripetal Force} = 	ext{Coulomb Force}$$\frac{mv^2}{r_0} = \frac{1}{4 \pi \varepsilon_0} \frac{e \cdot e}{r_0^2}$Simplifying the equation for $v^2$:$v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_0 m}$Taking the square root on both sides:$v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 r_0 m}} = \frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$Thus,the speed of the electron is $\frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$.

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