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In a radioactive material the activity a...

In a radioactive material the activity at time `t_(1)` is `R_(1)` and at a later time `t_(2)`, it is `R_(2)`. If the decay constant of the material is `lambda`, then

A

`R_(1) = R_(2)`

B

`R_(1) = R_(2) e^(-lambda (t_(1) - t_(2)))`

C

`R_(1) = R_(2) e^(lambda (t_(1) - t_(2)))`

D

`R_(1) = R_(2) (t_(2)//t_(1))`

Text Solution

Verified by Experts

The correct Answer is:
B
According to activity law, `R = R_(0) e^(-lambda t)`
`therefore" "R_(1) = R_(0) e^(-lambda t_(1)) and R_(2) = R_(0) e^(-lambda t_(2))`
`therefore" "(R_(1))/(R_(2)) = (R_(0)e^(-lambda t_(1)))/(R_(0) e^(-lambda t_(2))) = e^(-lambda t_(1)) e^(lambda t_(2)) = e^(0lambda(t_(1) - t_(2)))`
or, `R_(1) = R_(2) e^(-lambda (t_(1) - t_(2)))`
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