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(C) According to Bohr's theory,the angular momentum of an electron in an orbit is quantized as $mvr = \frac{nh}{2\pi}$.This implies that the orbital v

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$3$

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(C) According to Bohr's theory,the angular momentum of an electron in an orbit is quantized as $mvr = \frac{nh}{2\pi}$.This implies that the orbital velocity $v$ is inversely proportional to the radius $r$,i.e.,$v \propto \frac{1}{r}$.Let $v_1$ and $r_1$ be the velocity and radius in the ground state,and $v_2$ and $r_2$ be the velocity and radius in the excited state.Given $v_1 = v$,$r_1 = R$,and $v_2 = \frac{v}{3}$.Using the relation $\frac{v_1}{v_2} = \frac{r_2}{r_1}$,we get $\frac{v}{v/3} = \frac{r_2}{R}$.Solving for $r_2$,we find $3 = \frac{r_2}{R}$,which gives $r_2 = 3R$.

In a hydrogen atom,an electron excites from the ground state to a higher energy state,and its orbital velocity is reduced to $\frac{1}{3}$ of its initial value. The radius of the orbit in the ground state is $R$. The radius of the orbit in that higher energy state is: (in $R$)

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