The half-lives in two different samples, A and B, of radioactive nuclei are related according to T1/2,B = 1 6 T1/2,A. In a certain period the number of radioactive nuclei in sample A decreases to one-fourth the number present initially. In this same period the number of radioactive nuclei in sample B decreases to a fraction f of the number present initially. Find f.
(its not 1/3.. its not 0.917... and its not .875) Please answer to 4 decimal places
given that If Ta and Tb are half-lives of two samples,
let number of initial sample be Na and Nb respectively.
as they decay exponentially with the formula:
number of nuclei at any time t for A=Na*exp(-lambda*t)
where lambda=ln(2)/half life period
hence for A, if at time t , the nuclei sample has decreased to 1/4th of original numbers,
as lambda for B=ln(2)/Tb
hence number of samples left=Nb*exp(-lambda*t)=Nb*exp(-0.086643)=0.917*Nb
hence value of f is 0.917.
[note: please double check whether Tb=16*Ta or Tb=1.6*Ta in the question given to you. in the question uploaded, it is showing Tb=16*Ta and with that, the correct answer is f=0.9170.]
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