Question

Calculate the wavelength of the photons emitted by hydrogen atoms when they undergo transitions from n = 4 to n = 3.

Answer 1

Answer:

The hydrogen atom emitted light, meaning that that the electron must come from a higher energy level to a lower energy level. The wavelength of the light emitted is 434 nm.

Answer 2

The wavelength of the photons emitted by hydrogen atoms when they undergo transitions from $n=4$ to $n=3$ is $1875 nm$.

Explanation:

• Hydrogen is a single electron species.
• The electron is de-excited from fourth energy level to third energy level.
• We are to use Rydberg's formula to find out the wavelength of photon emitted.
• Rydberg's equation is $& \frac{1}{\lambda }=\text{R}\left( \frac{1}{\text{n}_{1}^{2}}-\frac{1}{\text{n}_{2}^{2}} \right)$ where 'n' represents the respective energy states and 'lambda' the wavelength of the photon..

• On substituting the values in Rydberg's equation and on solving we get,

$& \frac{1}{\lambda }=\text{R}\left( \frac{1}{\text{n}_{1}^{2}}-\frac{1}{\text{n}_{2}^{2}} \right) \\ \\\[\frac{1}{\lambda }=109677\times \left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\]$

$& =109677\times \left( \frac{1}{9}-\frac{1}{16} \right) \\ & =109677\times \left( \frac{7}{144} \right)$

$\[\lambda =18.75\times {{10}^{-5}}~\text{cm}=1875\times {{10}^{-9}}~\text{m}=1875~\text{nm}\]$