Question

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

Answer 1

given that If Ta and Tb are half-lives of two samples,

Tb=16*Ta...(1)

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,

Na/4=Na*exp(-lambda*t)

==>exp(-lambda*t)=1/4

==>t=1.3863/lambda

using lambda=ln(2)/Ta=0.69315/Ta

t=1.3863/(0.69315/Ta)=2*Ta....(2)

t=2*Ta=2*(1/16)*Tb=(1/8)*Tb

as lambda for B=ln(2)/Tb

==>t=(1/8)*ln(2)/lambda=0.086643/lambda

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.]