Question

Below you can see the energy levels of the Helium atom. The right axis is a quantum number related to angular momentum (do not worry too much about that). An electron is in the state 2s and after a little while it decays back to the ground state. What is the energy of the photon emitted?

Answer 1

ΔE = 20 eV

Step-by-step explanation:

In a Helium atom we have two electrons in the s layer, so they can accommodate one with the spin up and the other with the spin down, give us a total spin of zero (S = 0) this state is singlet, in general this very stable states,

When you transition to the 1s state to complete the two electrons allowed by layers

ΔE = -5 - (-25) = 20 eV

this is the energy of the transition,

It should be mentioned that there can also be transitions with the two spins of the same orientation, but in this case the energy is a little different due to the electron-electron repulsion, this state is called ortho helium S = 1