There is a stream of neutrons with a kinetic energy of `0.0327 eV`. If the half-life of neutrons is `700 s`, what fraction of neutrons will decay before they travel is distance of `10 m`? Given mass of neutron `= 1.676 xx 10^(-27) kg` .

Explanation

The velocity v of neutron is given by
`(1)/(2) m v^(2)=0.0327 eV`
or `(1)/(2) xx 1.675 xx 1^(0-27)v^(2)=0.0327 xx 1.6 xx 10^(-19) J`
`:. V^(2) = (2xx 0.0327 xx 1.6 xx 10^(-19))/((1).675 xx 10^(-27))`
or ` v=2.5 xx 10^(3) m s^(-1)`
Time taken by neutron to travel a distance of `10 m` is
`t=(distane)/(velocity) =(10)/(2.5 xx 10^(3)) =4 xx10^(-3) s`
and half -life `T =700s`.
The friction of neutrons decayed in time t is
`(N)/(N_(0)) = e^(-lambda t)=((1)/(2))^(t//T) =((1)/(2))^(4 xx 10^(-3)//700) =((1)/(2)) 5.7 xx 10^(-6)`
`=0.999952`
Fraction of neutrons decayed is
`1-0.999952 =0.000048 =4.8 xx10^(-5)` .