Gold (Au) has a synthetic isotope that is relatively unstable. After 25.5 minutes, a 128-gram sample has decayed to 2 g.
What is the half-life of this isotope?
8.5 minutes
2.25 minutes
1.5 minutes
4.25 minutes
x = Xi e^-kt
2 = 128 e^-25.5 k
0.015625 = e^-25.5 k
ln 0.015625 = -25.5 k
-4.15888 = -25.5 k
k = 0.163
so
1 = 2 e^-0.163 t for half life
0.5 = e^-0.163 t
ln 0.5 = -0.693 = -0.163 t
t = 4.25 minutes
128/2 = (1/2)^6
So, now you want
25.5/6 = 4.25
For a half-life of k minutes, your function is A = P*(1/2)^(t/k)
To find the half-life of the isotope, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t / T₁/₂)
Where:
N(t) is the amount of the substance at time t
N₀ is the initial amount of the substance
T₁/₂ is the half-life of the substance
t is the time that has passed
In this case, after 25.5 minutes (t = 25.5), the sample has decayed to 2 grams (N(t) = 2), and the initial amount of the sample is 128 grams (N₀ = 128).
2 = 128 * (1/2)^(25.5 / T₁/₂)
To solve for T₁/₂, we can rearrange the equation:
(1/2)^(25.5 / T₁/₂) = 2 / 128
Taking the logarithm of both sides:
(25.5 / T₁/₂) * log(1/2) = log(2 / 128)
Simplifying:
(25.5 / T₁/₂) * (-0.301) = -2.079
Dividing both sides by (-0.301):
(25.5 / T₁/₂) = 6.901
Now, solve for T₁/₂:
T₁/₂ = 25.5 / 6.901
T₁/₂ ≈ 3.70 minutes
Therefore, the half-life of this isotope is approximately 3.70 minutes.
None of the given options (8.5 minutes, 2.25 minutes, 1.5 minutes, 4.25 minutes) is the correct answer.
To determine the half-life of the isotope, we need to understand its radioactive decay. Given that the 128-gram sample has decayed to 2 grams after 25.5 minutes, we can use the formula for exponential decay:
N = N₀ * (1/2)^(t/T)
Where:
N = final amount
N₀ = initial amount
t = time elapsed
T = half-life
In this case, N = 2 g, N₀ = 128 g, and t = 25.5 minutes. We can plug these values into the formula and solve for T.
2 = 128 * (1/2)^(25.5/T)
To solve for T, divide both sides by 128:
2/128 = (1/2)^(25.5/T)
Convert the fraction on the left-hand side to a decimal:
0.015625 = (1/2)^(25.5/T)
We can rewrite the right-hand side as 2^(-25.5/T), since 1/2 is equal to 2^(-1):
0.015625 = 2^(-25.5/T)
To isolate the exponential term, take the logarithm of both sides (base 2):
log₂(0.015625) = log₂(2^(-25.5/T))
Use the logarithmic property that log₂(a^b) = b * log₂(a):
log₂(0.015625) = -25.5/T * log₂(2)
Simplify the right-hand side since log₂(2) equals 1:
log₂(0.015625) = -25.5/T * 1
Now, divide both sides by -25.5:
log₂(0.015625) / -25.5 = 1 / T
Calculate the left-hand side value:
log₂(0.015625) / -25.5 ≈ -0.943
Therefore, the reciprocal of -0.943 gives us the value of T:
T ≈ 1 / -0.943
T ≈ -1.06
Since negative time is not meaningful in this context, we take the absolute value:
T ≈ 1.06
Therefore, the half-life of this isotope is approximately 1.06 minutes.
Considering the provided answer choices, none of them match the calculated half-life.