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If `80%` of a radioactive element undergoing decay is left over after a certain periof of time `t` form the start, how many such periofs should elapse form the start for just over `50%` of the element to be left over? |
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Answer» Correct Answer - 3 Let `a=100, (a-x)=80` (Amount left). `:.` Amount left `=a/((2)^(n)), [n= "number of half lives"]` `80=100/((2)^(n))implies(2)^(n)=10/8` `:. n log 2= log 10-3 log 2 =1-3xx0.3=0.1` `n=0.1/(log 2)=0.1/0.3=1/3` `n=1/3=(t(("Time for 80% amount"),("left or 20% decomposed")))/(t_(1//2)(("Time for 50% amount"),("left or decomposed")))implies t_(1//2)=3t` Alternatively `t_(30%("left"))/t_(1//2("left"))=(log(100/80))/0.3=1/3` `t_(1//2)=3t_(80%("left"))=3t`Correct Answer - 3 Let `a=100, (a-x)=80` (Amount left). `:.` Amount left `=a/((2)^(n)), [n= "number of half lives"]` `80=100/((2)^(n))implies(2)^(n)=10/8` `:. n log 2= log 10-3 log 2 =1-3xx0.3=0.1` `n=0.1/(log 2)=0.1/0.3=1/3` `n=1/3=(t(("Time for 80% amount"),("left or 20% decomposed")))/(t_(1//2)(("Time for 50% amount"),("left or decomposed")))implies t_(1//2)=3t` Alternatively `t_(30%("left"))/t_(1//2("left"))=(log(100/80))/0.3=1/3` `t_(1//2)=3t_(80%("left"))=3t` |
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