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)

The radius of the \(n\)-th orbit in a hydrogen atom is proportional to the square of the principal quantum number \(n\), i.e., \(r_n \propto n^2\).

Given the ratio of the radii of the initial and final orbits:
\(\dfrac{r_i}{r_f} = \dfrac{16}{4}\)

This implies:
\(\dfrac{n_i^2}{n_f^2} = \dfrac{16}{4} \Rightarrow n_i = 4 \text{ and } n_f = 2\)

The wavelength \(\lambda\) of the emitted photon is given by the Rydberg formula:
\(\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)\)

Substituting the values:
\(\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right)\)

\(\dfrac{1}{\lambda} = 1.0973 \times 10^7 \left( \dfrac{1}{4} - \dfrac{1}{16} \right)\)

\(\dfrac{1}{\lambda} = 1.0973 \times 10^7 \times \dfrac{3}{16}\)

\(\lambda = \dfrac{16}{3 \times 1.0973 \times 10^7} \text{ m}\)

\(\lambda \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}\)

Answer: \(486\)