A radioactive substance decays at a rate of 30% per year. Presently there are 400 grams of the substance. How long before there are 10 grams?

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Answer 1 · 2016-05-28T16:54:42

#ln 40/0.3# years = 12.296 years = 12 years and 108 days, nearly.

If x gm is present at time t years,

# x'=-0.3x# gm/year. Integrating,

#int1/x dx=-0.3int dt#.

So, #ln x=-0.3t + A#. Inversely,

#x=c^(-0.3t+A)= e^Ae^(-0.3t)=Ce^(-0.3t)#

Initially, t = 0 and x = 400. So, C=400, and now,

#x = 400 e^(-0.3t)#

The time t years for decay, from 400 gm to 10 gm, is given by

#10=400e^(-0,3t)#. So, #e^(0.3t)=40#, and inversely,

#0.3t=ln 40#.

So, t=ln 40/0.3=12.296 years= 12 years and 108 days, nearly.