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After 3 days a sample of radon-222 has decayed to 58% of its original amount.

a) What is the half-life of radon-222?

Answers:

The number of radioactive nuclei,  at a time,  is related to the original number of nuclei,  at the time,  by, \( N =N_0 (\frac{1}{2})^n \)

=> \( N =N_0 (\frac{1}{2})^n \)

=> \( 58% N_0 =N_0 (\frac{1}{2})^n \) [convert persantage to number]

=> \( \frac{58}{100} N_0 =N_0 (\frac{1}{2})^n \)

=> \( 0.58 N_0 =N_0 (\frac{1}{2})^n \)

=> \( 0.58 \frac{N_0}{N_0}=(\frac{1}{2})^n \)

=> \( 0.58 =(\frac{1}{2})^n \)

=> \( \log 0.58 =n \log 0.5\)

Therefore => \( \frac{\log 0.58}{\log 0.5} =n \)

But \( n =\frac{t}{T_\frac{1}{2}} \)

=> \( T_\frac{1}{2}  =\frac{t}{n} \) [put the value of t and n]

=> \( T_\frac{1}{2}  =\frac{3}{\frac{\log 0.58}{\log 0.5}} \) [simplify]

=> \( T_\frac{1}{2}  = 3.8174 \; days \)

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