'Rn ' decays into 'Po' by emitting a - particle with half life of 4, days . A sample

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13.Nuclei

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$'Rn$' decays into $'Po'$ by emitting $a -$ particle with half life of $4\, days$. A sample contains $6.4 \times 10^{10}$ atoms of $Rn$. After $12\, days$, the number of atoms of $'Rn'$ left in the sample will be

A

$3.2 \times {10^{10}}$

B

$0.53 \times {10^{10}}$

C

$2.1 \times {10^{10}}$

D

$0.8 \times {10^{10}}$

Solution

$\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{t}{\mathrm{T}_{\mathrm{N}}}}$

$=6.4 \times 10^{10}\left(\frac{1}{2}\right)^{\frac{12}{4}}$

$=\frac{6.4}{2^{3}} \times 10^{10}=\frac{6.4}{8} \times 10^{10}$

$=0.8 \times 10^{10}$

Standard 12

Physics

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