The half life of P32 and P33 are 14 & 25days. These are mixed in ratio of 4:1of their atoms. If initial activity of mixed sample is 3 milicurie.find the activity after 60years.
(ans.=0.205mCi)

let X & Y denote the number of nuclie of x & y initially...

t_1/2 for x = 14 so (lamda)_x  = 0.693/14=0.0495 day^-1

t_1/2 for y = 25 so (lamda)_y = 0.028 day^-1

lamda is disinteration constant.......

initial activity = X(lamda)_x + Y(lamda)_y   ...........................1

X/Y = 4                                      ..........2                                                     (given)

from 1 & 2 

[   Y = 4.91*10⁸  &  X = 1.96*10⁹   ]                              

now

       number of nuclie at any time = N/N_i = (1/2)^t/t1/2

       from here number of nuclie remaining after 60 years for X & Y will be

X¹ = X(1/2)^60/14          &         Y¹ = Y(1/2)^60/25

rate of disintergration   = N(lamda)

 so net rate after 60 years = X¹(lamda)_x + Y¹(lamda)_y

                 =X(lamda)_x(1/2)^60/14 +  Y(lamda)_y(1/2)^60/25 

    now calculate net rate it will be nearly 2....