When an electron of energy 12.1 ev

Text Solution

`1.05 xx 10^(-34) J s``2.11 xx 10^(-34) J s``3.16 xx 10^(-34) J s``4.22 xx 10^(-34)`

Answer : B

Solution : Since `E = - (13.6)/(n^(2)) eV` <br> `E_(1) = - 13.6 eV` <br> `E_(2) = - 3.4 eV` <br> `E_(3) = - 15.0 eV` <br> `E_(4) = - 0.85 eV` <br> Force above, we can see that <br> `E_(3) - E_(3) = 12.1 eV` <br> i.e., the electron must be making a transition from `n = 3 to n = 1` level. <br> `Delta L = (3 - 1) (h)/(2 pi) = (h)/(pi) = 2.11 xx 10^(-34) J s`

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Here is part of the energy level diagram of hydrogen:

n=4 --> -0.85eV

n=3 --> -1.50eV

n=2 --> -3.40eV

n=1 --> -13.6eV

When an electron of energy 12.1eV collides with this atom, photons of three different energies are emitted. Show on the diagram (with arrows) the transitions responsible for these three photons.

I tried solving this problem by working out the difference of energies that would give 12.1eV. I noticed that -13.6-(-1.5)=12.1. However I wasn't able to progress from here and would appreciate any pointers/solutions. (Note: This isn't homework, it's from a worksheet I found online.)

Thanks