λ=100nm
E=hcλ=1242100=12.42eV
a) The possible transitions may be E1 to E2
E1 to E2, energy absorbed = 10.2 eV.
Energy left = 12.42 - 10.2 = 2.22 eV.
2.22 eV = hcλ=1242λ
or λ=559.45=560nm
E1 to E3, Energy absorbed = 12.1 eV
Energy left = 12.42 - 12.1 = 0.32 eV
0.32=hcλ=1242λ
or λ=12420.32=3881.2=3881nm
E3 to E4, Energy absorbed = 0.65
Energy left = 12.42-0.65 = 11.77eV
11.77=hcλ=1242λ
or λ=124211.77=105.52
b)The energy absorbed by the H atom is now radiated perpendicular to the incident beam.
→10.2=hcλ or λ=124210.2=121.76nm
→12.1=hcλ or λ=124212.1=102.64nm
→0.65=hcλ or λ=12420.65=1910.76nm