Question

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`.

Answer 1

Mass of neutron-mass of proton`= m`

From conseervation of momentum in y-direction,
`sqrt(2 mK_(1)) sin 45^(@) = sqrt(2 mK_(2)) sin theta` …(i)
In x-direction
`sqrt(2 mK_(0)) = sqrt(2 mK_(1)) cos 45^(@) + sqrt(2 mK_(2)) cos theta` …(ii)
Squarring and adding Eqs, (i) and (ii), we have
`K_(2) = K_(1) + K_(0) - sqrt (2K_(0)K_(1))` ...(iii)
From conseervation of energy ,
`K_(2) = K_(0) + K_(1)` [as collision is elastic] ...(iv)
Solving Eqs, (iii) and (iv), we get
`K_(1) = (K_(0))/(2)`
i.e. after each collision energy remain half. Therefore , after `n` collision ,
`K_(n) = K_(0) ((1)/(2))^(n)`
`:.0.23 = (4.6 xx 10^(6)) ((1)/(2))^(n)` `2^(n) = (4.6 xx 10^(6))/(0.23)`
Taking log and solving, we get
`n = 24`