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In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to $\frac{1}{3}$ rd of its initial value. The radius of the orbit in the ground state is $\mathrm{R}$.
The radius of the orbit in that higher energy state is
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The radius of the orbit in that higher energy state is
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The correct answer is:
$3 \mathrm{R}$
According to Bohr's orbit
$m v r=\frac{n h}{2 \pi}$
i.e., $\quad \mathrm{v} \propto \frac{1}{\mathrm{r}}, \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}}$
$\frac{v}{v / z}=\frac{r_{2}}{R} \Rightarrow r_{2}=3 R$
$m v r=\frac{n h}{2 \pi}$
i.e., $\quad \mathrm{v} \propto \frac{1}{\mathrm{r}}, \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}}$
$\frac{v}{v / z}=\frac{r_{2}}{R} \Rightarrow r_{2}=3 R$
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