Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer 1

`4.86*10^(-7)\ \text{m}`

Step-by-step explanation:

R = Rydberg constant =
`1.09677583* 10^7\ \text{m}^(-1)`

`n_1` = Principal quantum number of an energy level = 2

`n_2` = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

`\lambda^(-1)=R\left((1)/(n_1^2)-(1)/(n_2^2)\right)\\\Rightarrow \lambda^(-1)=1.09677583* 10^7\left((1)/(2^2)-(1)/(4^2)\right)\\\Rightarrow \lambda=\left(1.09677583* 10^7\left((1)/(2^2)-(1)/(4^2)\right)\right)^(-1)\\\Rightarrow \lambda=4.86*10^(-7)\ \text{m}`

The wavelength of the light emitted is
`4.86*10^(-7)\ \text{m}`.