A neutron with an energy of `4.6 MeV` collides elastically with proton and is retarded. Assuming that upon each collision the neutron is deflected by `45^(@)`, find the number of collisions which will reduce its energy to `0.23 eV`.

Explanation

Mass of neutron`~~` mass of proton `=m`

From conservation of momentum in y-direction
`sqrt(2mK_(1)) sin 45^(@)=sqrt(2mK_(2)) sin theta`.....(i)
In x-direction `sqrt(2mK_(0))-sqrt(2mK_(1))cos 45^(@)=sqrt(2mK_(2))cos theta`.....(ii)
Squaring and adding equation (i) and (ii), we have
`K_(2)=K_(1)+K_(0)-sqrt(2K_(0)K_(1))`......(iii)
From conservation of energy
`K_(2)=K_(0)-K_(1)`.......(iv)
Solving equations (iii) and (iv), we get
`K_(1)=(K_(0))/(2)`
i.e., after each collision energy remains half. Therefore, after `n` collisions,
`K_(n)=K_(0)((1)/(2))^(n)`
`:. 0.23-(4.6xx10^(6))((1)/(2))^(n) 2^(n)=(4.6xx10^(6))/(0.23)`
Taking log and solving, we get
`n ~~ 24`