The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of 0.084 eV, travells a distance of 1 km? (mass of neutron `1.68 xx 10^(-27)` kg, and In 2=0.693)

Options

A. `60 xx 10^(-5)`
B. `15 xx 10^(-5)`
C. `25 xx 10^(-5)`
D. `50 xx 10^(-5)`

Explanation

Given , the half - life of neutron , `t_(1//2) = 693 sec`
Kinetic energy of neutron = 0.084 eV
`1/2 mv^2 = 0.084 xx 1.6 xx 10^(-19)`
`v^2 = (2 xx 0.084 xx 1.6 xx 10^(-19) )/(m)`
` = (2 xx 0.084 xx 1.6 xx 10^(-19) )/(1.68 xx 10^(-27) ) = 0.16 xx 10^8`
` v = 0.4 xx 10^4 m//s `
Time taken by neutron to travelled in 1 km,
` t = (1000)/(0.4 xx 10^4) rArr t = 0.25 s`
By radioactive decay's law ,
`N = N_0 (1/2)^n rArr (N)/(N_0) = (1/2)^n`
`(N_0)/(N) = 2^n`
Taking log on the both sides , we get
` ln (N_0)/(N) = ln 2^n = n ln 2 = (t)/(t_(1//2) ) ln 2 = (0.5 xx 0.693 )/(693)`
` = 0.25 xx 10^(-5) " " [ because ln 2 = 0.693 ]`