Question
Question: The count rate from \(100\text{ c}{{\text{m}}^{3}}\) of a radioactive liquid is \(c\). Some of this ...
The count rate from 100 cm3 of a radioactive liquid is c. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid in cm3, is
A.20
B.40
C.60
D.80
Solution
The first step should be to find out a relation between the count rate and the volume of the liquid. Then three half-lives mean, the count rate is halved after each half life, which means that the count rate of the remaining liquid decreases by a factor of 8 eventually.
Complete answer:
It is given that the count rate from 100 cm3 of a radioactive liquid is c, then count rate of c/10 will be for a volume of 10 cm3.
As it is given that some of the liquid is discarded from a total volume of 100 cm3, then let us assume that the volume of discarded liquid is x cm3. If an amount x is withdrawn, then the remaining amount will be:
100−x cm3
As it is given that the count rate of the remaining liquid is found to be c/10 after three half-lives, i.e., the volume of the remaining liquid becomes 10 cm3 after three half-lives. We also know that after each life the liquid will be halved and then eventually becomes 10 cm3, therefore equating these, we get:
2×2×2100−x=10⇒8100−x=10⇒100−x=8×10∴100−x=80
Here 100−x is the volume of the remaining liquid when x cm3 of the liquid is withdrawn from the initial volume. Therefore, the volume of the remaining liquid would be 80 cm3.
Hence the correct option is D out of the given options.
Note:
One more way to solve this question was first find out the count rate for 1 cm3 of the liquid, which will be c/100. After three half lives the count rate would become 81×100c and now if the volume of the remaining liquid is taken as V, then we can write the equation as V×81×100c=10c.