1.

A radioactive sample S_1 having the activity A_1, has twice the number of nuclei as another sample S_2 of activity A_2." If "A_2= 2A_1, then the ratio of half-life of S_1 to the half-life of S_2 is :

Answer»

4
2
0.25
0.75

Solution :Activity `A=lambdaN =(0.693)/(T_(1//2))N`
where `T_(1//2)` is the half-life of a radiactive SAMPLE.
`A_(i)/A_(2)=(N_(1))/T_(1) xx T_(2)/N_(2)`
`T_(1)/_(2)=A_(2)/A_(1) xx N_(1)/N_(2)`
`=(2A_(1))/(A_(1)) xx (2N_(2))/(N_(2))=4/1`


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