Question

An electron in an unknown energy level of a hydrogen atom transitions to the n=2 level and emits a photon with wavelength 410 nm in the process. What was the initial energy level? Use R[infinity]=2.179×10−18J for the hydrogen atom Rydberg constant. Use h=6.626×10−34 Js for Planck's constant. Use c=2.998×108ms for the speed of light. Your answer should be a whole number.

Answer 1

6

Step-by-step explanation:

`n_1` = Final energy level = 2

`R_H` = Rydberg constant =
`10967758.3\ \text{m}^(-1)`

`\lambda` = Wavelength = 410 nm

From the Rydberg formula we have

`(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2))\\\Rightarrow n_2=((1)/(n_1^2)-(1)/(\lambda R_H))^{-(1)/(2)}\\\Rightarrow n_2=((1)/(2^2)-(1)/(410* 10^(-9)* 10967758.3))^{-(1)/(2)}\\\Rightarrow n_2=\pm6.01\approx 6`

The initial energy level is 6.