When an excited electron in a hydrogen atom falls from =5 to =2, a photon of blue light is emitted. if an excited electron in an he+ ion falls from =4, which energy level must it fall to ) for blue light of a similar wavelength to be emitted?

n = 3

Explanation:

Given the formula for the transition energy of an atom with 1 electron:

For the H transition n=5 to n=2:

Then we solve for nf with Z=2 (Helium)

Is near 3, actually the energy of the transitions are:

H (5⇒2) = -2.85 eV = 434 nm (Dark blue)

He (4⇒3) = -2.64 eV = 469 nm (Light blue)

I thought it was cool to see the actual colors. Included them.

3

explanation: