Question

Suppose an e- in an H atom has a transition from n = 3 to n = 2.
a. Determine the energy of the released photon.

i set it up like this
Ei=-Rh/3^2 Ef=-Rh/2^2

(-Rh/9)-(-rh/4)=-4rh+9rh/36=/-rh+4rh/9/- this part why does it go to -rh+4rh? But then I got the end result which is 3rh/9 =hv

Then I did 3/9 x 2.179x10^-18 j/6.626 x 10^-34 j.s =1.10 x 10^17/s

b. determine what type of light is emitted.
Show all units and conversion factors.

For this part I did 3.00x10^8 m/s/1.10x10^-6m =273 nam

Is this correct?

Answer 1

I follow some of what you did but not all of it and I don't get the same answer you did. Why not use
dE = 2.180E-18J(1/4-1/9)
Then lambda = hc/dE
For dE I have about 3.03E-19J
and wavelength = about 6.56E-7 m or 656 nm or 6560 angstroms which is red.