1.

A and B are the two radioactive elements. The mixture of these elements show a total activity of 1200 disintegrations/minute. The half-life of A is 1 day and that of B is 2 days. What will be the total activity after 4 days ? Given, the initial number of atom in A and B are equal :

Answer»

200 dis/min
250/min
500 dis/min
150 dis/min

Solution :We have `A=lambda(N) =(0.693)/(T_(1//2))(N)`
Initial number of atoms is A and B are same
`A_(0) alpha 1/T_(1//2)`
`rArr A_(0)/A_(0) ((A))/((B)) =(48 hr)/(24 hr)=2`
ALSO `A_(0) (A)+A_(0) (B)=1200`
`rArr 3A_(0) (B)=1200`
`A_(0) (B)=400`
and `A_(0) (A)=800`
So, `A(A)=(A_(0)(A))/(2^(4))=(800)/(16)=50`
`and A(B)=(A_(0)(B))/((2)^(2))=(400)/(4)=100`
Hence total activity after 4 days is
`=A_(0) (A)+A_(0)(B)=50+100=150"dis/min"`


Discussion

No Comment Found

Related InterviewSolutions