Suppose that initially you have a 100 grams of a certain radioactive isotope. After 15 years, there are only 10 grams left of the isotope. Find an equation for Q(t), the amount left of the isotope in grams, after t-years. What is the half life of this isotope?
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Q =Qo e^(-lambda x t)
You need to evaluate the decay constant lambda
re-arranging gives:-
Q/Qo=e^(-lambda x t)
taking natural logs of both sides to get rid of e.
ln (Q/Qo) =- lamda xt
Q = 10; Qo = 100; t = 15
1/t x ln(Q/Qo) = -lambda
remove - sign on rhs by inverting ln on lhs
lambda = 1/t x ln(Qo/Q)
lamda = 1/15 x ln (100/10)
lambda = ln (10)/15
Half life, t, is when Q = 1/2 x Qo
or, Q/Q0 =1/2
Q/Qo=e^(-lambda x t)
1/2 = e^(-lambda x t)
ln (1/2) = -lambda x t
1/lambda x ln (1/2) = -t
we get rid of - sign on rhs bt inverting log on lhs to give
1/lambda x ln (2/1) =t
t= ln (2)/lambda
t = (ln (2) ln (10))/15
t = (15 x ln(2))/ln(10)
You need to evaluate the decay constant lambda
re-arranging gives:-
Q/Qo=e^(-lambda x t)
taking natural logs of both sides to get rid of e.
ln (Q/Qo) =- lamda xt
Q = 10; Qo = 100; t = 15
1/t x ln(Q/Qo) = -lambda
remove - sign on rhs by inverting ln on lhs
lambda = 1/t x ln(Qo/Q)
lamda = 1/15 x ln (100/10)
lambda = ln (10)/15
Half life, t, is when Q = 1/2 x Qo
or, Q/Q0 =1/2
Q/Qo=e^(-lambda x t)
1/2 = e^(-lambda x t)
ln (1/2) = -lambda x t
1/lambda x ln (1/2) = -t
we get rid of - sign on rhs bt inverting log on lhs to give
1/lambda x ln (2/1) =t
t= ln (2)/lambda
t = (ln (2) ln (10))/15
t = (15 x ln(2))/ln(10)
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Q(t) = 100e^(-at)
0.1 = e^(-15a)
a = ln(10)/15
Once you get the decay constant a, half life T equals ln(2)/a
T = 15ln(2)/ln(10)
0.1 = e^(-15a)
a = ln(10)/15
Once you get the decay constant a, half life T equals ln(2)/a
T = 15ln(2)/ln(10)