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+0 Answers


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

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

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