For a stream of neutrons with a kinetic energy of 0. 335 e V, the fraction of neutrons that will decay before they travel a distance of 32 m is found to be `N xx 10^(- 5)` . Find the value of N if the half-life of neutrons is given to be 700 s. (Take the mass of a neutron =` 1. 675 xx 10^(- 27)` kg and `2^(- 5.7xx10^(- 6)` = 0. 999952)

Answer

4.8

Explanation

The K.E. of neutron is given by,
`1/2mv^2= 0.335 eV`
`therefore` `1/2mv^2 = 0. 335 xx 1. 6 xx 10^(-19)` J
`therefore v^2 = (2 xx 0.335 xx 1.6 xx 10^(-19))/(1.675 xx 10^(-27))` = `64 xx 10^6`
implies `v = 8 xx 10^3 ms^(-1)`
Time taken by neutron to travel a distance of 32 m,
`t = ("distance")/("velocity") = (32)/(8 xx 10^1) = 4 xx 10^(-3)s`
Half-life, T = 700 s
The fraction of neutrons remaining after time t ,
`N/(N_0) = e^(-lamdat)`
`therefore` `N/(N_0)` = `(1/2)^(1/r) = (1/2) ^(4 xx 10^(-3))/(700)`
= `(1/2)^(5.7 xx 10^(-6)`
= 0.999952