Question
Question: An electron is lying initially in the n = 4 excited state. The electron de-excites itself to go to n...
An electron is lying initially in the n = 4 excited state. The electron de-excites itself to go to n = 1 state directly emitting a photon of frequency n41.If the same electron first de-excites to n = 3 state by emitting a photon of frequency n43 and then goes from n = 3 to n = 1 state by emitting a photon of frequency n31, then
A
n41 = n 43 + n 31
B
n 41 = n 43 – n31
C
n 43= n 41 + 2n31
D
Data Insufficient
Answer
n41 = n 43 + n 31
Explanation
Solution
E = E1 + E2