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The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. `14 mi n`B. `20 mi n`C. `28 mi n`D. `7mi n` |
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Answer» Correct Answer - B Here, `T=20 "minutes"` In time `t_(1)`, the radioactive sample left behind `=1-1/3=2/3`. In time `t_(2)`, the radioactive sample left behind `=1-1/3=2/3`. As `n/(N_(0))=(1/2)^(t//T) :. 2/3(1/2)^(t_(1)//T).........(i)` and `1/3=(1/2)^(t_(2)//T)` Dividing (ii) by (i), we get `1/2=(1/2)^((t_(2)-t_(1))//T)` or `t_(2)-t_(1)=T=20min`. |
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