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Calculate the half -life period of a radioactive element that remains only \frac{1}{64} of its original amount in 4500 yrs?

Option: 1

1185 yrs


Option: 2

 750 yrs


Option: 3

850 yrs


Option: 4

700 yrs


Answers (1)

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\mathrm{\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}} where \mathrm{n=N_{0}} of half live.

\mathrm{\therefore \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\frac{1}{64} \quad \because N=\frac{1}{64} N_{0}}
\mathrm{n= 6}

\mathrm{\text { Total time } =n \times t_{1 / 2} }
                          \mathrm{ =6 \times t_{1 / 2}=4500 }
\mathrm{ t_{1 / 2} =750 \mathrm{yrs} }
                                  

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