Register to reply 
Halflife of radioactive substance 
Share this thread: 
#1
Feb811, 12:45 PM

P: 86

1. The problem statement, all variables and given/known data
If 20% of a radioactive substance disappears in 70 days, what is its halflife? 2. Relevant equations y = C*e^(k*t) where t is time in days k is the constant of proportionality? y is the current amount of substance 3. The attempt at a solution 20% disappears, so 100%  20% = 80% remaining = 0.80 after 70 days .8*C = C*e^(70*k) Solve for k k = 3.188 * 10^3 Plug k value back into equation to find t at the substance's halflife (.5) .5*C = C*e^((3.188 * 10^3)*t) Solve for t t = 217.44 217.44 days is the substance's halflife. Is that correct? Thank you. 


#2
Feb811, 02:38 PM

P: 343

It matches what I get.



Register to reply 
Related Discussions  
Half Life of radioactive element  High Energy, Nuclear, Particle Physics  6  
Half Life of radioactive element  Introductory Physics Homework  1  
Half Life of Radioactive Isotopes  High Energy, Nuclear, Particle Physics  1  
Half Life of radioactive insulin  Chemistry  1  
Finding the halflife of an unknown substance  Introductory Physics Homework  6 