there is a strea of neyrons with a kinetic energy of 0.0327 eV if the half - life of neutrions be 700 s , what fraction of neutrons will decay before they travel a distance of 10m? Mass of neutron equal to ` 1.67xx10^(-27) kg`

Answer

`0.004`

Explanation

Given that kinetic energy of neutrons is
`(1)/(2) mv^(2)=0.0327 xx(1.6 xx10^(-19)) J`
or `v^(2)=(2 xx 0.0327 xx(1.6 xx10^(-19)))/(1.675 xx10^(-27)) =625 xx10^(4)`
or `v=2500 ms^(-1)`
Time to travel a distance of 10 km is
`(10^(4) (m))/(2500 m s^(-1))=4s`
After `4s`, number of neutrons left can be given us
`N =N_(0) 2^(-n)`
where `n=(t)/(T)= no`. of half- lives. Here, `n=(4)/(700) =(1)/(175)`
or `(N)/(N_(0)) =2^(-1//175)=0.996` or `N=0.996 N_0`
Thus, fraction of neutrons decayed are
`f=(N_(0)-N)/(N_(0)) =(0.004N_(0))/(N_(0)) =0.004`