Q136 / 180 · KCET 2012 · Physics

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

  • A
    $9 \mathrm{R}$
  • B
    $2 \mathrm{R}$
  • Correct answer
    C
    $3 \mathrm{R}$
  • D
    $27 \mathrm{R}$

Solution

According to Bohr's orbit $m v r=\frac{n h}{2 \pi}$ i.e., $\quad \mathrm{v} \propto \frac{1}{\mathrm{r}}, \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}}$ $\frac{v}{v / z}=\frac{r_{2}}{R} \Rightarrow r_{2}=3 R$

Hence, the correct answer is option: 'C'.

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