A sample of a radioactive substances undergo 80% decomposition in 345 minutes its half life is
Answers
Answered by
5
The answer for this ques is ln2/ln5×345
Answered by
20
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
Attachments:
Similar questions