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