Given φ=2.25eV λ=300×100−3m Energy of the given photon, E=hv =
hc
2
=
663×10−34×3×108
300×10−9
=6.63×10−19J=4.13eV Now, E=
hc
2
−φ=4.137−2.25=1.88eV→(1) So, 1.88eV energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen ⇒n=1234 E=−13.6ev&−3.4&−1.51&−0.85 And from statement (1), 1.18=Ei−Ef {Ei&Ef are energy of initial and final orbit\} And also from the above table, we can observe that, Ei−Ef=(−1.51)−(−3.4)=1.89eV =E2−E3 That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 300nm wavelength hydrogen electron can jump from 3 to 2 .