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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?

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]

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