Consider a point sorce emiiting `alpha`- particles and receptor and receptor of area `1 cm^(2)` placed `1m` away from source. Receptor records any `alpha`-particle falling on it. If the contains `N_(0) = 3.0 xx 10^(16)` active nuclie and the receptor records a rate of `A = 50000 "counts"//"second"`. Assume that the source emits alpha particles unifomrly in all direction and the alpha particles fall nearly normally on the window. If decay constant is `3n xx 10^(-(n+1))`, then the value of `n`

Answer

`7`

Explanation

Let `lambda=` decay constant
`N_(0)=` number of active nuclei `=3.0xx10^(16)`
number of `alpha`-particles emitted per second `=lambdaN_(0)`
number of `alpha`-particle falling on the window `=(lambdaN_(0)a)/(4piR^(2))`
`lambda=(4piR^(2)A)/(N_(0)a)=(4pixx5xx10^(4))/(3.0xx10^(16)xx10^(-4))=2.1xx10^(-7)s^(-1)=(7(3))/(10)xx10^(-7)=(7xx3)10^(-8)=3xx7xx10^(-(7+1))`