A sample of a radioactive substance undergoes 80 decomposition in 345 minutes
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N=No e^-lamba(345)
20=100 e^-lamba(345)
1/5=e^-lamba(345)
ln1/5=-lamba(345)
Lamba=ln5 /345 ......(1)
T(1/2)=ln 2/lamba
Lamba =ln2/T(1/2).........(2)
Substituting lamba value
ln5/345=ln2/T(1/2)
T(1/2)=ln 2/ln5 x 345
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