The correct option is B I → Q II → P,S,U; III → P,R,S,U; IV → P,R,T,U;
K.E for perfectly inelastic (head on) collision ⇒
m1V0=(m1+m2)V [By COM] ...... (1)
12m1V20=12(m1+m2)v2+Q [By COE] ...... (2)
By (1) and (2) K=Q(1+m1m2) here K=12m1V20
and Q is energy of photon emitted in various transition.
For Li,
Q=13.6×32(1−1n2) n = 1, 2, 3, ....
Q=122.4(1−1n2)⇒Q1=91.8,Q2=108.8,Q3=114.75
Qmax=122.4
Also, K=Q(1+m1m2)=Q(1+17)=Q87
Kmin=Q1×87=104.9eV
Kmax=Qmax×87=122.4×87=139.8eV
Question I:
K=81.6
K<Kmin. So only elastic collision is possible.
Question II:
K=104.9
K=Kmin. So perfectly inelastic collision can happen, and electron can get excited. If electron is not excited then elastic collision will happen.
Question III:
K=108.8
So, perfectly inelastic collision can happen by exciting the electron to 2nd level. Or, inelastic collision can happen by exciting the electron to the 1st level.
Question IV:
K=200>Kmax
Inelastic collision can happen exciting the electron. K>Kmax so, the atom can get ionised.