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`        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)`
7 years ago

510 Points
```										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....
```
7 years ago
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