As the amount of radioactive isotope remaining after 2 hrs is 1/16 of the intial amount,the eauation N=(N)initial x (exp(-kt)), here k denote lambda, takes the form N)initial/16=N)initial x (exp(-kt)) i.e., exp(-kt)=1/16 => -kt= log(1/16) .....[taking log on both the sides] =>-kt=-4log2 now time taken is 2 hrs therfore, 2k=4log2 => k=2log2 now half life =log2/k=1/2hrs=30 mins ,hence the answer (ii) THIS WAS FOR NCERT TRADITIONAL WORMS NEXT SOLUTION IS A BETTER APPROACH the amount left after n-half lives=initial amount/(2 raised to power n),now as the final amount 1/16 ,i.e., 1/(2 raised to power 4) of the initial amount, 4 half lives have passed therefore, 4(half life)=2 hrs => half life=30 min
118) After two hours 1/16th of the initial amount of a certain radioactive isotope remains undecayed. The half life of the isotope is
Answer is:
30 min
Explanation:
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