Q173 / 180 · COMEDK 2017 · Physics

The electron in a hydrogen atom makes a transition from $n=n_{1}$ to $n=n_{2}$ state. The time period of the electron in the initial state $n_{1}$ is eight times that in the final state $n_{2}$. The possible values of $n_{1}$ and $n_{2}$ are

  • A
    $n_{1}=8, n_{2}=1$
  • Correct answer
    B
    $n_{1}=4, n_{2}=2$
  • C
    $n_{1}=2, n_{2}=4$
  • D
    $n_{1}=1, n_{2}=8$

Solution

Here, for hydrogen atom, $Z=1$ We know that, time period of electron in $n$th orbit is given as $\begin{aligned} T_{n} & \propto n^{3} \\ \therefore \quad & \frac{T_{n_{1}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \end{aligned}$ Since, $\quad T_{n_{1}}=8 T_{n_{2}}$ $\therefore \quad \frac{8 T_{n_{2}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \Rightarrow 2^{3}=\left(\frac{n_{1}}{n_{2}}\right)^{3}$ $\Rightarrow \quad n_{1}=2 n_{2}$ Hence, this condition satisfies only in option (b). i.e., $\quad n_{1}=4$ and $n_{2}=2$

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

Relevant Questions