The expression for the radius of the n^{th} B | vedclass

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(B) $1$. According to Bohr's model, the centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^

Vedclass · Accepted Answer

$r = \frac{n^2 h^2}{4 \pi^2 m e^2 Z}$

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(B) $1$. According to Bohr's model, the centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$.$2$. Bohr's quantization condition is $mvr = \frac{nh}{2\pi}$, which gives $v = \frac{nh}{2\pi mr}$.$3$. Substituting $v$ into the force equation: $\frac{m}{r} (\frac{nh}{2\pi mr})^2 = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$.$4$. Simplifying for $r$: $r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$.$5$. In $CGS$ units, where $\frac{1}{4\pi\epsilon_0} = 1$, the expression becomes $r = \frac{n^2 h^2}{4\pi^2 m Z e^2}$, which matches option $(B)$.

The expression for the radius of the $n^{th}$ Bohr orbit of a hydrogen-like atom is:

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