The half-life period of a radioactive element A is 62 years. It decays into another stable element B. An…

The half-life period of a radioactive element A is 62 years. It decays into another stable element B. An archeologist found a sample in which A and B are in 1:15 ratio. The age of the sample is
  1. 248 years
  2. 186 years
  3. 124 years
  4. 310 years

Solution

$\begin{aligned} & \text { } \frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{15} \Rightarrow \frac{\mathrm{~N}}{\mathrm{~N}_0-\mathrm{N}}=\frac{1}{15} \\ & \Rightarrow 15 \mathrm{~N}=\mathrm{N}_0-\mathrm{N} \Rightarrow \frac{\mathrm{N}}{\mathrm{N}_0}=\frac{1}{16} \\ & \text { Also, } \frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{62}} \\ & \Rightarrow \frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{62}} \Rightarrow\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{62}} \\ & \therefore \mathrm{t}=62 \times 4=248 \text { years }\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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