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Half -life for a radioactive isotope of Indium is 69300\, \mathrm{sec}.The time required to fall the rate of decay to \left(\frac{1}{10}\right)^{\text {th }} of its initial value is = ?

Option: 1

2.303 \times 10^{5} \mathrm{~s}


Option: 2

4.606 \times 10^{5} s


Option: 3

2.303 \times 10^{4} s


Option: 4

4.6 \times 10^{4}s


Answers (1)

best_answer

\mathrm{\lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{69300}=10^{-5} \mathrm{~s}^{-1}}
\mathrm{t=\frac{1}{\lambda} \ln \left(\frac{R_{0}}{R}\right)=\frac{1}{10^{-5}} \ln (10)}

\mathrm{ t=2.303 \times 10^{5} s=230300 \, s}

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sudhir.kumar

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