 # 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)

12 years ago

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

t1/2 for x = 14 so (lamda)x  = 0.693/14=0.0495 day-1

t1/2 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/Ni = (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....