There is a stream of neutrons with kinetic energy of 0.0327 eV. If half life of neutrons is 700s, what fraction of neutrons will decay before they travel a distance of 10m? Take mass of neutron =`1.675xx10^(-27)kg.`

Explanation

Kinetic energy of neutrons is given to be 0.0327 eV ,From the relation ,` K=(1)/(2) mv^(2)` , we can find the speed of the neutorns
`therefore k=(1)/(2) mv^(2)`
Or `v=sqrt((2K)/(m))=sqrt((2xx0.0327xx1.6xx10^(-19))/(1.675xx10^(_27)))`
`=2.5xx10^(-3)ms^(-1)`
time taken by the neutrons to travel a disance of 10 m with this speed ,`t=(10)/(2.5xx10^(3))=4.0xx10^(-3)S`
fraction of neutrons decayed in the given time interval ,
`=(N_(o)(1-e^(-lamdat)))/(N_(0))=(1-e^(-lamdat))`
where ,` t_(1//2)=700s`
`t_(1//2)("half -life")=(0.693)/(lamda ), (lamda = "decay constant") `
` implies lamda =0.693//t_(1//2)=(0.693)/(700)`
` =1-e^(-((0.693)/(700))(4.0xx10^(-3)))=3.96xx10^(-6)`