What is the half-life in years of the substance described ?

Question

The mass m of a substance at time t is modelled by the equation m = m0e^-kt where m0 is the initial mass and k is a constant. After 20 years, 1/5 has decayed and 4/5 remain.

Answer 1

Niall,

You've got 2 different answers and you need to know which is correct. Well, it was the first one.

Sub in m = (4/5) m0 and t = 20 and what you get (in its simplest form) is:

m = m0 (4/5)^(t/20)

Then, sub in m = m0 / 2 to find the time taken for half the substance to decay. After taking logs this gives:

t = 20ln(1/2) / ln(4/5)

which is 62.13 years (4sf)

Hope this helps


Perspy


PS GeekCreole rounded his k to 3sf before using that to calculate an answer to 5sf! That's where he went wrong, I guess.

PPS to AngelHeart. I never questioned GreekCreole's procedure. However, rounding to 3sf and using this result to derive a result which you quote to 5sf is wrong and that is why his answer is wrong. Similarly, you have rounded k to 4sf and used your result to derive an answer which you quote to 4sf. It is therefore WRONG in the 4th sf.

Answer 2

Quantities that are subject to exponential decay are commonly denoted by the symbol N. (This convention suggests a decaying number of discrete items. This interpretation is valid in many, but not all, cases of exponential decay.) If the quantity is denoted by the symbol N, the value of N at a time t is given by the formula:


where N0 is the initial value of N (at t = 0)

When t = 0, the exponential is equal to 1, and N(t) is equal to N0. As t approaches infinity, the exponential approaches zero. In particular, there is a time such that


Substituting into the formula above, we have

Answer 3

Your question is not totally clear. Half life usually refers to radioisotopes or biological or environmental t1/2.

They all have unique functions with regards to mass.

In the argument of mass alone, a body of 20Kg with a half life of 5 years will be 10Kg after 5 yrs, 2.5Kg after 10 yrs, 1.25kg after 15 years and so on.

Try playing with exponents.

Hope this helps.

Answer 4

Geek Creole made a CLEAR procedure
k =
ln (4/5)
______
20

k = .01116 which could round to k = .0112

Geek Creole Procedure is right on Clear
ln (1/2) = kt
t =
ln(1/2)
______
k
or t = .69315/k
t = .69315/ .01116

t = 62.10996
or t = 62.11 years

Answer 5

.8m0 = m0e^(-20k)
.8 = e^-20k
ln(.8) = -20k
k = ln(.8)/-20 = .011157
m= m0e^(-.011157t)
.5m0 = m0e^-.011157t
.5 = e^-.011157t
ln(.5) = -.011157t
t= ln(.5)/(-.011157) = 62.13 years

Answer 6

First solve for k
ln(m/m0)=-kt
ln(4/5)=-k20
0.0112=k

Now solve for t when m/m0=1/2
ln(1/2)=-0.0112t
61.8881yrs = t