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 is 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)

For hydrogen atom:

$$radius\ r_n\propto n^2$$

given:

$$r_i:r_f=16:4=4:1$$

so:

$$n_i^2:n_f^2=4:1\Rightarrow n_i:n_f=2:1$$

 choose quantum numbers

Since electron moves from higher to lower orbit:

nᵢ > n_f

Smallest integers satisfying ratio 2:1 are:

$$n_i​=4,n_f​=2$$

use Rydberg formula:

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

$$=1.0973\times10^7\left(\frac{1}{4}-\frac{1}{16}\right)$$

$$=1.0973\times10^7\cdot\frac{3}{16}\approx2.057\times10^6$$

$$lambda=\frac{1}{2.057\times10^6}\approx4.86\times10^{-7}m$$ 

=486 nm