Question
In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits in given by $r_{i}:r_{f} = 16:4$. The wavelength of photon emitted due to this transition is $\_\_\_\_$ nm. (Given Rydberg constant $= 1.0973 \times 10^{7}/m$ )
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Solution
Sol. [Key Concept: $r \propto \ n^{2}$ gives $n_{i} = 4$, $n_{f} = 2$; then $\frac{1}{\lambda} = R(\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$.]
Given : $\frac{r_{i}}{r_{f}} = \frac{16}{4} = \frac{n_{1}^{2}}{n_{f}^{2}}$
$$\begin{matrix} & \frac{n_{i}}{n_{f}} = \frac{4}{2} \\ & \frac{1}{\lambda} = R\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right) = 1.0973 \times 10^{7}\left( \frac{1}{(2)^{2}} - \frac{1}{(4)^{2}} \right) \\ & \lambda = 4.86 \times 10^{- 7}\text{ }m \approx 486\text{ }nm \end{matrix}$$
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