If t_{1/2} is the half-life of a substance,th | vedclass

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(A) The half-life $t_{1/2}$ is defined as the time required for a radioactive substance to decay to half of its initial amount.Similarly,$t_{3/4}$ rep

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Decays $\frac{3}{4}^{th}$

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(A) The half-life $t_{1/2}$ is defined as the time required for a radioactive substance to decay to half of its initial amount.Similarly,$t_{3/4}$ represents the time required for the substance to decay to $\frac{3}{4}$ of its initial amount.According to the law of radioactive decay,the amount remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $t_{1/2}$,$N = \frac{N_0}{2}$,which gives $t_{1/2} = \frac{\ln(2)}{\lambda}$.For $t_{3/4}$,the amount decayed is $\frac{3}{4}N_0$,so the amount remaining is $N = N_0 - \frac{3}{4}N_0 = \frac{1}{4}N_0$.Using $N = N_0 e^{-\lambda t_{3/4}}$,we get $\frac{1}{4} = e^{-\lambda t_{3/4}}$,which implies $4 = e^{\lambda t_{3/4}}$.Taking the natural logarithm on both sides,$\ln(4) = \lambda t_{3/4}$,so $2\ln(2) = \lambda t_{3/4}$.Since $\lambda = \frac{\ln(2)}{t_{1/2}}$,we have $2\ln(2) = \frac{\ln(2)}{t_{1/2}} 	imes t_{3/4}$,which simplifies to $t_{3/4} = 2t_{1/2}$.Thus,$t_{3/4}$ is the time in which the substance decays $\frac{3}{4}^{th}$ of its initial value.

If $t_{1/2}$ is the half-life of a substance,then $t_{3/4}$ is the time in which the substance:

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