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. `60xx10^(-5)`
B. `15 xx10^(-5)`
C. `25xx10^(-5)`
D. `50xx10^(-5)`

Answer

C

Explanation

Given,
Half life of neutron, `t _(1//2) =693s`
Kinetic energy of neutron, K.E `- 0.084eV`
`= 0.084 xx 1.6 xx 10 ^(-19) V`
Distance travelled by neutron, `d =1 km = 1000 m`
Mass of neutroo n `, =1.6 xx 10 ^(-27) kg`
`ln 2 =0.693`
According to Radioative decay law,
`N=N_(0) ((1)/(2)) ^(n )`
` (N_(0))/(N ) =2^(@)`
ln `(N_(0))/(N) = ln 2 ^(n) (because` Applying log on both sides)
`ln"" (N_(0))/(N) = n ln 2`
`= n( 0.693)`
`= (((d )/(v ))/(t _(1//2)))0.693 (because V = (d)/(t ))`
` = (100)/(693 sqrt((2xx 0.084 xx 1.6 xx 10 ^(-13))/(1.68 xx 10 ^(-27))))xx 0.693`
`= sqrt((1.68 xx 10 ^(-27))/(0.2688xx 10 ^(-19)))`
`= sqrt(6.25 xx 10 ^(8))`
` =2.5 xx 10 ^(4)`
`therefore ln "" (N _(0))/(N) = 0/25 xx 10 ^(-5)`