A sample of radioactive element containing 4 × 10^{16} active nuclei. Half life of element is 10 da

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13.Nuclei

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A sample of radioactive element containing $4 \times 10^{16}$ active nuclei. Half life of element is $10$ days, then number of decayed nuclei after $30$ days is ........ $\times 10^{16}$

A

$0.5$

B

$2$

C

$3.5$

D

$1$

(AIPMT-2002)

Solution

No. of active nuclei present in original sample $=4 \times 10^{16}$

Half-life of element $=10$ days.

No. of half-life in 30 days $=\frac{30}{10}=3$

No. of nuclei present after 30 days $=\frac{4 \times 10^{16}}{(2)^{3}} \Rightarrow 0.5 \times 10^{16}$

No. of decayed nuclei = (No. of active nuclei present originally) – (No. of Active nuclei present after $30$ days)

$=4 \times 10^{16}-0.5 \times 10^{16}$

$3.5 \times 10^{16}$

Standard 12

Physics

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