InterviewSolution
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InterviewSolution
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20% of a radioactive substances decay in 10 days . Calculate the amount of the original material left after 30 days.A. `78%`B. `62%`C. `51%`D. `48%` |
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Answer» Correct Answer - C ( c) amount of element left undecayed after 10 days `=1-(20)/(100)=(8)/(10)` now ,`N =(8)/(10)N_(0) and t= 10` days we have ,`N=N_(0) e^(-lamda t)` `(8)/( 10) N_(0)=N_(0) e^(-lamda t) implies =(10)/(8)implies lamda =(log_(e) (10//8))/(10)` ` therefore lamda = 0.022` if `T_(1//2)` is half -life period ` T_(1//2)=(0.693)/(lamda ) =(0.693)/(0.022)=31.5 `days Again ,` N=N_(0) ((1)/(2))^(n)=N_(0) ((1)/(2))^((31.5)/(30))` `implies % "of" N//N_(0) = 51.2%` |
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