MHT CET · Physics · Nuclear Physics
A sample of radioactive element contains \(8 \times 10^{16}\) active nuclei. The halt-life of the element is 15 days. The number of nuclei decayed after 60 days is
- A \(7.5 \times 10^{+16}\)
- B \(2.0 \times 10^{16}\)
- C \(0.5 \times 10^{16}\)
- D \(4.0 \times 10^{16}\)
Answer & Solution
Correct Answer
(A) \(7.5 \times 10^{+16}\)
Step-by-step Solution
Detailed explanation
Half life, \(\mathrm{T}=15\) days, time \(\mathrm{t}=60\) days \(=4 \mathrm{~T}\) The number of nuclei remaining is given by
\(
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{N}_0\left(\frac{1}{2}\right)^4=\frac{1}{16} \times 8 \times 10^{16} \\
& =0.5 \times 10^{16} \\
& \therefore \text { No. of nuclei decayed }=\mathrm{N}_0-\mathrm{N} \\
& =8 \times 10^{16}-0.5 \times 10^{16}=7.5 \times 10^{16}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}=\mathrm{N}_0\left(\frac{1}{2}\right)^4=\frac{1}{16} \times 8 \times 10^{16} \\
& =0.5 \times 10^{16} \\
& \therefore \text { No. of nuclei decayed }=\mathrm{N}_0-\mathrm{N} \\
& =8 \times 10^{16}-0.5 \times 10^{16}=7.5 \times 10^{16}
\end{aligned}
\)
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