Question

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =3 to =1.

Answer 1

`1.936* 10^(-18)\ \text{J}`

Step-by-step explanation:

`R_h` = Rydberg constant =
`2.178* 10^(-18)\ \text{J}`

`n_i` = Initial shell = 3

`n_f` = Final shell = 1

We have the relation

`\Delta E=R_h((1)/(n_f^2)-(1)/(n_i^2))\\\Rightarrow \Delta E=2.178* 10^(-18)((1)/(1^2)-(1)/(3^2))\\\Rightarrow \Delta E=1.936* 10^(-18)\ \text{J}`

The energy of the photon emitted here is
`1.936* 10^(-18)\ \text{J}`.