+0  
 
0
600
3
avatar+216 

Iodine-131 is a radioactive form of iodine. After the crisis at a Japanese nuclear power plant in March 2011, elevated levels of this substance were detected thousands of miles away from Japan. Iodine-131 has a half-life of 8 days. What is the daily decay factor for this substance? (Round your answer to two decimal places.) _______

idenny  Mar 14, 2017
 #2
avatar+88871 
0

Call A the original amount of the substance....and we have this equation

 

.5A = Ae^(-k*t)     where k is the decay factor and t is the time in days = 8

........divide both sides by A

 

.5 =  e^(-k*8)     take the ln of both sides

 

ln (.5)   = ln e^(-k*8)     and we can write

 

ln (.5)  =( -k*8)       and ln e   = 1   so we can ignore this

 

ln(.5)  = -k * 8        divide both sides by -8

 

-ln(.5) / 8  = k ≈ 0.09 [rounded ]

 

 

cool cool cool

CPhill  Mar 14, 2017
 #3
avatar
0

Constant of decay, when half-life is known, is always:

ln(1/2) / half-life =ln(1/2) / 8 =-0.08664......[constant of decay is always negative].

Guest Mar 14, 2017

9 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.