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

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 ___________.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

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