The half life of a certain radioactive substance is 10 years. How long will it take for 18 gms of the substance to decay to 6 gms?
A. 6 ln 10 years
B. 10 ln 6 years
C. 10 (ln 3)/(ln 6) years
D. 18 ln 10 years
E. 10 (ln 3)/(ln 2) years
.5 = e^-k(10)
-.693 = -10 k
k = .0693
then
6/18 = e^-.0693 t
-1.0986 = - .0693 t
t = 15.9 years
I would solve
18(1/2)^(t/10) = 6
(1/2)^(t/10) = 1/3
from the answers, my clue would be to take ln of both sides
Thanks I think it would be E.10 (ln3)/(ln2) years
Right?
10 ln 3/ln 2 = 15.9 yes
P = P(0)e^(kt)
(1/2) = e^(10k)
ln(1/2) = 10k
k = (1/10)ln(1/2)
----------------
6 = 18e^((1/10)ln(1/2)t)
(1/3) = e^((1/10)ln(1/2)t)
t = 10[ln(1/3)/ln(1/2)]
t = 10[-ln(3)/-ln(2)]
t = 10[ln(3)/ln(2)]
To find out how long it will take for the substance to decay from 18 grams to 6 grams, we can use the equation for exponential decay:
N(t) = N₀ * (1/2)^(t/t₁/₂)
where:
N(t) = the amount of substance remaining at time t
N₀ = the initial amount of substance (18 gms in this case)
t = the time elapsed
t₁/₂ = the half-life of the substance (10 years in this case)
We can rearrange the equation to solve for t:
N(t)/N₀ = (1/2)^(t/t₁/₂)
6/18 = (1/2)^(t/10)
1/3 = (1/2)^(t/10)
To solve this equation, we can take the logarithm (base 2) of both sides:
log₂(1/3) = log₂[(1/2)^(t/10)]
- log₂(3) = (t/10) * log₂(1/2)
- log₂(3) = (t/10) * (-1)
log₂(3) = t/10
t = 10 * log₂(3)
To find this value, we can use the properties of logarithms:
log₂(3) = ln(3)/ln(2)
So, t = 10 * ln(3)/ln(2)
Therefore, the correct answer is E. 10 (ln 3)/(ln 2) years.