SAMPLE PROBLEM (FOR BETTER UNDERSTANDING)

1. Consider strontium-90 which has a half-life of approximately 28 years.
   • Initially, at time t=0, the sample is 100% strontium-90
   • After 28 years, only half the original amount of strontium will remain:
         ½ x 100% = 50%
   • After another 28 years, only half of this amount of strontium-90 will remain:
         ½ x 50% = 25%
   • After another 28 years, only half of this amount of strontium will remain:
         ½ x 25% = 12.5%
   • and so on.
   • At any given time,
         the amount of strontium-90 that has undergone decay can be calculated:     amount of strontium-90 decayed = the original amount - the amount remaining.
   
   

   Number of Half-lives Time (years) % Strontium-90 remaining % Strontium-90 that has decayed 0 0 100 0 1 28 50 50 2 56 25 75 3 84 12.5 87.5 4 112 6.25 93.75 5 140 3.125 96.875 6 168 1.5625 98.4375

2.  Calculate the percentage of strontium-90 remaining after 280 years.
3. N_t = N_o x (0.5)^{number of half-lives}
   N_t = ? %
   N_o = 100%
   number of half-lives = time ÷ half-life = 280 ÷ 28 =10
   N_t = 100 x (0.5)¹⁰ = 0.098%
   
   Reference: (http://www.ausetute.com.au/halflife.html)