Question

What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answer 1

`\lambda=1282nm`

Step-by-step explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

`(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})`

Here
`R_H` is the Rydberg constant for hydrogen and
`n_1,n_2` are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for
`\lambda`

`(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\\(1)/(\lambda)=7.81*10^5m^(-1)\\\lambda=(1)/(7.81*10^5m^(-1))\\\lambda=1.282*10^(-6)m\\\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\\\lambda=1282nm`