KCET · Chemistry · Chemical Kinetics
\(2 \mathrm{~g}\) of a radioactive sample having half-life of 15 days was synthesised on 1st Jan 2009. The amount of the sample left behind on 1st March, 2009 (including both the days) is
- A \(0 \mathrm{~g}\)
- B \(0.125 \mathrm{~g}\)
- C \(1 \mathrm{~g}\)
- D \(0.5 \mathrm{~g}\)
Answer & Solution
Correct Answer
(B) \(0.125 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{llrl}\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} & & \\ \text { Given, } & \mathrm{N}_{0} =2 \mathrm{~g} \\ \mathrm{t}_{1 / 2} =15 \text { days } \\ \Rightarrow & \mathrm{T} =60 \text { days } \\ & \mathrm{n} =\frac{60}{15}=4 \\ & \mathrm{~N} =2\left(\frac{1}{2}\right)^{4} \\ \text { or } & \mathrm{N} =0.125 \mathrm{~g}\end{array}\)
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