OBronium-12 is a radioactive material that spontaneously decays when it comes in contact with oxygen. One minute after a 200-mg sample of OBronium-12 is placed into an oxygen chamber, only 140 mg remains. Round your answers to 1 decimal place.
a) Determine the half-life of OBronium-12 in seconds.
b) After how long would there be 20% of OBronium-12 left in seconds?
it has a half-life, so we expect a function like
y = a(1/2)^(x/h)
where h is the half-life. We have the point (1,140) on the curve, so
200(1/2)^(1/h) = 140
1/2^(1/h) = 0.7
1/h = log0.7/log0.5 = 0.515
h = 1.943 minutes
and that means y = 200(1/2)^(x/1.943)
so now find x when y=0.2*200 = 40
1/2^(x/1.943) = 0.2
x/1.943 = log0.2/log0.5 = 2.322 (half-lives)
x = 4.512 minutes
oops. I just noticed that it wants the equations in terms of seconds, so just make that adjustment.
np, thank you so much
To determine the half-life of OBronium-12, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t / T)
Where:
- N(t) is the remaining amount at time t
- N₀ is the initial amount
- t is the time passed
- T is the half-life
We are given that after 1 minute, 140 mg of OBronium-12 remains from an initial 200 mg. Let's convert the time to seconds for consistency:
1 minute = 60 seconds
Now we can plug the values into the formula and solve for T:
140 = 200 * (1/2)^(60 / T)
Divide both sides by 200:
0.7 = (1/2)^(60 / T)
To eliminate the exponent, we can take the logarithm (base 0.5) of both sides:
log₀.₅(0.7) = (60 / T)
Using logarithm properties, we can rewrite the equation as:
T = 60 / log₀.₅(0.7)
Using a scientific calculator or an online logarithm calculator, we find:
T ≈ 143.7 seconds
a) The half-life of OBronium-12 is approximately 143.7 seconds.
Now, to determine the time after which only 20% of OBronium-12 remains, we can again use the exponential decay formula:
0.2 = 1 * (1/2)^(t / T)
Divide both sides by 1:
0.2 = (1/2)^(t / T)
Taking the logarithm (base 0.5) of both sides:
log₀.₅(0.2) = t / T
Rearranging the equation:
t = log₀.₅(0.2) * T
Again, using a scientific calculator or an online logarithm calculator, we find:
t ≈ 0.736 * 143.7
t ≈ 105.7 seconds
b) After approximately 105.7 seconds, there would be 20% of OBronium-12 remaining.