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  • The half-life of the radioactive radon is 3.8 days. The time, at the end of which <img alt="1 / 20^{\text {th }}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20/%2020%5E%7B%5Ctext%20%7Bt

The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1 / 20^{\text {th }} of the radon sample will remain undecayed, is (given \mathrm{\log _{10} \mathrm{e}=0.4343} )
 

Option: 1

\mathrm{3.8 \: days}

 


Option: 2

\mathrm{16.5 \: days}


Option: 3

\mathrm{33 \: days}


Option: 4

\mathrm{76 \: days}


Answers (1)

best_answer

Using \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}

where \mathrm{\lambda=\frac{\ln 2}{\mathrm{t}_{1 / 2}}=\frac{\ln (2)}{3.8}}

\mathrm{\therefore \frac{\mathrm{N}_0}{20}=\mathrm{N}_0 \mathrm{e}^{-\frac{\ln (2)}{3.8} \mathrm{t}}}
Solving this equation with the given data we find \mathrm{\mathrm{t}=16.5 \: days.}

Hence option 2 is correct.

Posted by

Nehul

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