How many head-on, eleastic coillisions must a neutron have with deuterium nucleus to reduce its energy from `1 MeV` to `0.025 eV`.

Explanation

Let mass of neutron `= m` and mass of deuterium `= 2m`
Initial kinetic energy of neutron `= K_(0)`
Let after first collision kinetic energy of neutron and deuterium be `K_(1)` and `K_(2)`.
Using `C.O.L.M` along direction of motion
`sqrt(2mK_(0))=sqrt(2mK_(1))+sqrt(4mK_(2))`
velocity of seperation `=` velocity of approach
`(sqrt(4mK_(2)))/(2m)-(sqrt(2mK_(1)))/(m)=(sqrt(2mK_(0)))/(m)`
Solving equation (i) and (ii) we get
`K_(1)=(K_(0))/(9)`....(1)
Loss in kinetic energy after first collision
`DeltaK_(1)=K_(0)-K_(1)`
`DeltaK_(1)=(8)/(9)K_(0)`....(1)
After second collision
`DeltaK=DeltaK_(1)=(8)/(9).(K_(0))/(9)`
`:.` Total energy loss
`DeltaK=DeltaK_(1)+......+DeltaK_(n)`
As, `DeltaK=(8)/(9)K_(0)+(8)/(9^(2))K_(0)+......+(8)/(9^(n))K_(0)`
`DeltaK=(8)/(9)K_(0)(1+(1)/(9)+.....+(1)/(9^(n-1)))`
`(DeltaK)/(K_(0))(8)/(9)[(1-(1)/(9^(n)))/(1-(1)/(9))]=1-(1)/(9^(n))`
Here, `K_(0)=10^(6)eV, DeltaK=(10^(6)-0.025)eV`
`:. (1)/(9^(n))=(K_(0)-DeltaK)/(K_(0))=(0.025)/(10^(6))` or `g^(n)=4xx10^(7)`
Taking log both sides and solving, we get
`n=8`