Question

A radioactive material is reduced to `1/16` of its original amount in 4 days. How much material should one being with so that `4xx10^-3kg` of the material is left over after 6 days ?

Answer 1

Here, `N/(N_0)=1/16 t=4days`
As `N/(N_0)=(1/2)^n=1/16=(1/2)^4 :. n=4`
Half life `T=t/n=4/4=1day.`
Now, `N=4xx10^-3kg, t=6days, N_0=?`
As `N/(N_0)=(1/2)^n=(1/2)^(6//1)=1/64`
`:. N_(0)= 64N=64xx4xx10^-3=0.256kg`