Q165 / 180 · KCET 2010 · Physics
An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transition. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transition. The ratio of the velocities of the electron in these two orbits is
- A$\frac{1}{2}$
- B$\frac{2}{1}$
- C$\frac{5}{4}$
- Correct answerD$\frac{3}{4}$
Solution
Number of spectral lines obtained due to transition of electrons from $n$th orbit to lower orbit is,
$\begin{array}{ll} & \mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2} \\ \text { I case } & 6=\frac{\mathrm{n}_{1}\left(\mathrm{n}_{1}-1\right)}{2} \\ \Rightarrow & \mathrm{n}_{1}=4 \\ \text { II case } & 3=\frac{\mathrm{n}_{2}\left(\mathrm{n}_{2}-1\right)}{2} \\ \Rightarrow & \mathrm{n}_{2}=3\end{array}$
Velocity of electron in hydrogen atom in $n$th orbit
$$
\begin{aligned}
\mathrm{v}_{\mathrm{n}} & \propto \frac{1}{\mathrm{n}} \\
\frac{\mathrm{v}_{\mathrm{n}}}{\mathrm{v}_{\mathrm{n}}} &=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \\
\Rightarrow \quad \frac{\mathrm{v}_{6}}{\mathrm{v}_{3}} &=\frac{3}{4}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.