The half life of a stream of radioactive particles moving along as straight path with a constant kinetic energy of 4 eV is 1 minute. The percentage of particles which decay before travelling a distance of 3.6 km is (Mass of the radioactive particles ` = 3 . 2 xx 10 ^(-21)` kg and charge of the electron ` = 1.6 xx 10 ^(-19) C` ).

Options

A. 87 . 5
B. 175
C. 37 . 5
D. 75

Answer

A

Explanation

Given , KE = 4eV `e = 1.6 xx 10^(-19) C`
`KE = (1)/(2) mv^(2)`
`rArr v^(2) = (2 xxKE)/(m)`
`= (2 xx 4 xx 1.6xx 10^(-19))/(32 xx 10^(-21)) = (2xx40)/(2 xx 10^(-2))`
`v^(2) = 400`
`rArr v = 20 m//s`
Time to travel a distance of 3.6 km is
`t = (D)/(v) = (3.6 xx 1000)/(20) s = (3.6xx1000)/(20 xx 60) min `
= 3 min
= 3 half -time
In three half life times the number of particles decayed
`~~ 87 . 5%`
`= (N_(0) - N)/(N_(0)) xx 100 , "where " N = (N_(0))/(8)`
`= (7)/(8) xx 100 = 87 . 5%`