A sample consists of nuclei of type A and nuclei of type B. Type A nuclei are radioactive and decay into type B nuclei B nuclei which are stable. The half-life for the radioactive decay A→B is T. At t = 0, the sample contains A1 nuclei of type A and B1 nuclei of type B such that A1B1=12. If at t=t0, the sample contains A2 nuclei of type A and B2 nuclei of type B such that A2B2=15, the ratio t0T is equal to ___________.

At t=t0, the number of nuclei of type A, A2=A1et0Tloge2At t=t0, the number of nuclei of type B, B2=B1+A1(1−et0Tloge2)Let et0Tloge2=pDividing the two equations, A2B2=A1pB1+A1(1−p)⇒ A2B2=A1B1p1+A1B1(1−p) ⇒ 15=p21+12(1−p) ⇒ p=12 ⇒ t0T=1

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A sample consists of nuclei of type A and nuclei of type B. Type A nuclei are radioactive and decay into type B nuclei B nuclei which are stable. The half-life for the radioactive decay $A \to B$ is T. At t = 0, the sample contains $A_{1}$ nuclei of type A and $B_{1}$ nuclei of type B such that $\frac{A_{1}}{B_{1}} = \frac{1}{2} .$ If at $t = t_{0},$ the sample contains $A_{2}$ nuclei of type A and $B_{2}$ nuclei of type B such that $\frac{A_{2}}{B_{2}} = \frac{1}{5},$ the ratio $\frac{t_{0}}{T}$ is equal to ___________.

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Detailed Solution

At $t = t_{0},$ the number of nuclei of type $A, A_{2} = A_{1} e^{\frac{t_{0}}{T} \log_{e} 2}$

At $t = t_{0},$ the number of nuclei of type $B, B_{2} = B_{1} + A_{1} (1 - e^{\frac{t_{0}}{T} \log_{e} 2})$

Let $e^{\frac{t_{0}}{T} \log_{e} 2} = p$

Dividing the two equations, $\frac{A_{2}}{B_{2}} = \frac{A_{1} p}{B_{1} + A_{1} (1 - p)}$

$\Rightarrow \frac{A_{2}}{B_{2}} = \frac{\frac{A_{1}}{B_{1}} p}{1 + \frac{A_{1}}{B_{1}} (1 - p)} \Rightarrow \frac{1}{5} = \frac{\frac{p}{2}}{1 + \frac{1}{2} (1 - p)} \Rightarrow p = \frac{1}{2} \Rightarrow \frac{t_{0}}{T} = 1$