Wavelength of Emitted Photon in Hydrogen Atom Transition

To find the wavelength of the emitted photon during the electronic transition in the hydrogen atom, we start from the expression for the radius of the $n^{\text{th}}$ orbit in Bohr's model:

$r_n \propto n^2$

Thus, the radii of the initial orbit $r_i$ and final orbit $r_f$ are related to their respective principal quantum numbers $n_i$ and $n_f$ by:

$\frac{r_i}{r_f} = \left(\frac{n_i}{n_f}\right)^2$

Given that the ratio of the radii is $r_i : r_f = 16 : 4$, we can write:

$\frac{n_i^2}{n_f^2} = \frac{16}{4}$

Since $n_i$ and $n_f$ are positive integers, this directly implies: $n_i = 4 \quad \text{and} \quad n_f = 2$

Now, using the Rydberg formula for hydrogen, the wavelength $\lambda$ of the emitted photon during the transition from $n_i = 4$ to $n_f = 2$ is given by:

$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

where $R = 1.0973 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant.

Substituting the values of $n_i$ and $n_f$:

$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$

$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} R$

Solving for $\lambda$:

$\lambda = \frac{16}{3 R}$

Substitute $R = 1.0973 \times 10^7 \text{ m}^{-1}$:

$\lambda = \frac{16}{3 \times 1.0973 \times 10^7 \text{ m}^{-1}} = \frac{16}{3.2919 \times 10^7} \text{ m}$

$\lambda \approx 4.8604 \times 10^{-7} \text{ m} = 486 \text{ nm}$

Hence, the wavelength of the emitted photon is 486 nm, which corresponds to Option C.