The radioactivity of a sample is R(1) at...

The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at time `T_(2)`. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time `(T_(2) -T_(1))` is proporational to

`(R_(1)T_(1) - R_(2)T_(2))`

`(R_(1)-R_(2))T`

`(R_(1) - R_(2))//T`

`(R_(1) - R_(2))(T_(1) - T_(2))`

Text Solution

Verified by Experts

The correct Answer is:

`N_(1)=N_(0)e^(-lambdaT_(1)), N_(2)=N_(0)e^(-lambdaT_(2))`
`R_(1)=lambdaN_(1), R_(2)= lambdaN_(2)`
`(N_(1)-N_(2))=lambda/lambda (N_(1)-N_(2))=((R_(1)-R_(2)))/(lambda)`
`T=(log_(e)2)/(lambda), lambda=(log_(e)2)/(T)`
`(N_(1)-N_(2))=((R_(1)-R_(2))T)/((log_(e)2))`
`(N_(1)-N_(2)) prop (R_(1)-R_(2)) T`

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The radioactivity of sample is R_(1) at a time T_(1) and R_(2) at a time T_(2) . If the half -life of thespeciman is T, the number of atoms that have disintegrated in the time (T_(2)-T_(1)) is equal to :

`(R_(1)T_(1)-R_(2)T_(2))`

`(R_(1)-R_(2))`

`(R_(1)-R_(2))//T`

`(R_(1)-R_(2))T//0.693`

The radioactivity of a sample is R_(1) at a time T_(1) and R_(2) at a time, T_(2) If the half-line of the secimen is T , the number of atoms that have disintergrated in the time (T_(2) - T_(1)) is proportional to:

`(R_(1) T_(1) - R_(2)T_(2))`

`(R_(2) - R_(1))`

`(R_(2) - R_(1))//T`

`(R_(2) - R_(1))T//0.693`

The radioactivity of a sample is l_(1) at a time t_(1) and l_(2) time t_(2) . If the half-life of the sample is tau_(1//2) , then the number of nuclei that have disintegrated in the time t_(2)-t_(1) is proportional to

`(l_(1)-t_(2)-l_(2)t_(1)`

`l_(1)-l_(2)`

`(l_(1)-l_(2))/(tau_(1//2))`

`(l_(1)-l_(2))tau_(1//2)`