The half-life of radioactive Radon is $3.8$ days. In how much time $(1 / 20) t h$ of the Radon sample will remain undecayed ? (Given: ${log}_{10} e = 0.4343$ )

13.8

days

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16.5

days

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33

days

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76

days

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Solution

The correct option is A $16.5$ days
${(\frac{1}{2})}^{n} = \frac{N}{N_{0}} = \frac{1}{20}$

$\Rightarrow n = 4.32$
$t = n T = 4.3 \times 3.8 = 16.5$ days

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of Radon sample will remain undecayed.

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The half-life of radioactive Radon is 3.8 days. In how much time (1/20)th of the Radon sample will remain undecayed ? (Given: log10e=0.4343)

The correct option is A 16.5 days(12)n=NN0=120 ⇒n=4.32t=nT=4.3×3.8=16.5 days

The half-life of radioactive Radon is $3.8$ days. In how much time $(1 / 20) t h$ of the Radon sample will remain undecayed ? (Given: ${log}_{10} e = 0.4343$ )

The correct option is A $16.5$ days
${(\frac{1}{2})}^{n} = \frac{N}{N_{0}} = \frac{1}{20}$

$\Rightarrow n = 4.32$
$t = n T = 4.3 \times 3.8 = 16.5$ days