# Question #7e7e4

Nov 10, 2016

The frequency of a photon emitted or absorbed in that electronic transition is $6.9 \times {10}^{14} \textcolor{w h i t e}{i} H z$.

#### Explanation:

The wavelength of the photon involved is given by the Rydberg's formula:

$\frac{1}{\lambda} = {R}_{H} \left(\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right)$,

where ${R}_{H}$ is the known Rydberg's constant, ${R}_{H} = 1.097 \times {10}^{7}$ ${m}^{- 1}$ and the numbers ${n}_{1}$ and ${n}_{2}$ are the initial and final first quantic numbers of energy levels of the electron.

Then we have:

$\frac{1}{\lambda} = 1.097 \times {10}^{7}$ ${m}^{- 1} \cdot \left(\frac{1}{2} ^ 2 - \frac{1}{5} ^ 2\right) = 2.304 \times {10}^{6}$ ${m}^{- 1}$.

The radiation frequency can be obtained from the equation of the propagation velocity in the vacuum:

$c = \lambda \cdot \nu$ $\Rightarrow$ $\nu = \frac{c}{\lambda} = c \cdot \frac{1}{\lambda}$

Then:

$\nu = 3 \times {10}^{8}$ $m \cdot {s}^{- 1} \cdot 2.304 \times {10}^{6}$ ${m}^{- 1} = 6.9 \times {10}^{14}$ $H z$.