Half-life calculus. Please help me understand
Favorites|Homepage
Subscriptions | sitemap
HOME > > Half-life calculus. Please help me understand

Half-life calculus. Please help me understand

[From: ] [author: ] [Date: 12-10-07] [Hit: ]
after t-years. What is the half life of this isotope?taking natural logs of both sides to get rid of e.Half life, t,or,......
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?

-
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)

-
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)
1
keywords: help,me,life,calculus,understand,Please,Half,Half-life calculus. Please help me understand
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .