._11^24Na(half-life=15hrs.)is known to contain some radioactive impurity (half-life=3hrs.) in a sample. This sample has an intial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. what percent of the intial activity was due to the impurity ?

._11^24Na(half-life=15hrs.)is known to contain some radioactive impuri

1.	._11^24Na(half-life=15hrs.)is known to contain some radioactive impurity (half-life=3hrs.) in a sample. This sample has an intial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. what percent of the intial activity was due to the impurity ?
Answer» 10 40 5 20 Solution :Let the activity due to IMPURITY be 'a' cpm `:.` due to NA it is (1000-a) cpm. After 30 hrs 'a' would be REDUCED to `(1/2)^10` a cpm and (1000-a) would be reduced to `1/4` (1000-a)cpm `:.` TOTAL activity after 30 hrs would be `(10/2)^10 a+1/4(1000-a)=200`(given) SOLVING we get `250-1/4a=200` `therefore 1/4a=50 implies a=200` Hence 20% activity was due to impurity.