किसी रेडियोऐक्टिव पदार्थ की समय `t_(1)`पर सक्रियता `R_(1)` तथा समय `t_(2)`है, पर सक्रियता `R_(2)` हो, तो औसत आयु का मान ज्ञात कीजिए।
(दिया है `t_(2) gt t_(1)`)

Explanation

हम जानते है कि , `R = R_(0) e^(-lambda t)`
` R_(1) = R_(0)e^(-lambda t_(1))`
`R_(2) = R_(0)e^(-lambda t_(2))" "{{:(.:. t = t_(1) " पर ," R = R_(1)),(.:. t = t_(2)" पर, "R = R_(2)):}}`
`:. R_(1)/R_(2) = (e^(-lambda t_(1)))/(e^(-lambda t_(2))) = e^(-lambdat_(1) + lambdat_(2)) = e^(lambda(t_(2) - t_(1)))`
या `log_(e) . R_(1)/R_(2) = lambda(t_(2) - t_(1)) " या " lambda = (log_(e) . R_(1)/R_(2))/(t_(2) - t_(1))`
औसत आयु `T = 1/lambda = (t_(2) - t_(1))/(log_(e). R_(1)/R_(2)) = (t_(2) - t_(1))/("In" R_(1)/R_(2))`