Use a calculator to help solve the problem.
An isotope of lead, 201Pb, has a half-life of 8.4 hours. How many hours ago was there 40% more of the substance? (Round your answer to one decimal place.)
____hr
To solve this problem, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t / T)
Where:
N(t) is the quantity of the substance at time t
N₀ is the initial quantity of the substance
T is the half-life of the substance
t is the time passed since the initial quantity
Let's denote the initial quantity as N₀ and the time passed as t. We are given that the isotope has a half-life of 8.4 hours, so we have T = 8.4 hours.
According to the problem, we want to find the number of hours ago when the quantity was 40% more than the current quantity. We can express this as:
N₀ + (0.4 * N₀) = N(t)
Simplifying this equation, we get:
1.4 * N₀ = N(t)
Now, let's substitute N(t) into the exponential decay equation and solve for t:
1.4 * N₀ = N₀ * (1/2)^(t / 8.4)
Canceling out N₀, we have:
1.4 = (1/2)^(t / 8.4)
Taking the logarithm of both sides, we get:
log₁/₂(1.4) = t / 8.4
Using a calculator to evaluate log₁/₂(1.4), we find:
t / 8.4 ≈ 0.494
Multiplying both sides of the equation by 8.4, we get:
t ≈ 4.15
Therefore, the isotope of lead, 201Pb, was 40% more than the current quantity approximately 4.15 hours ago.
To solve this problem, we can use the formula for exponential decay:
N(t) = N0 * (1/2)^(t/T)
Where:
N(t) is the amount of substance at time t,
N0 is the initial amount of substance,
t is the time that has passed,
T is the half-life of the substance.
In this case, we know the half-life (T) is 8.4 hours. We want to find the time (t) when there was 40% more of the substance, which means N(t) = 1.4 * N0 (since 1 + 0.4 = 1.4).
Substituting these values into the equation, we have:
1.4 * N0 = N0 * (1/2)^(t/8.4)
Dividing both sides of the equation by N0, we get:
1.4 = (1/2)^(t/8.4)
To solve for t, we need to take the logarithm of both sides of the equation. In this case, we can use the logarithm base 2 to match with the base (1/2) in the equation:
log2(1.4) = log2((1/2)^(t/8.4))
Using logarithmic property, we can bring down the exponent:
log2(1.4) = (t/8.4) * log2(1/2)
Now, we can solve for t by dividing both sides of the equation by log2(1/2):
t/8.4 = log2(1.4) / log2(1/2)
To find the value of log2(1.4) / log2(1/2), you can use a scientific calculator or an online calculator that supports logarithmic functions.
Calculating this expression, we find:
t/8.4 ≈ ~0.6736
Multiplying both sides of the equation by 8.4 to isolate t, we have:
t ≈ ~5.6574
Rounded to one decimal place, the answer is approximately 5.7 hours.
Therefore, approximately 5.7 hours ago, there was 40% more of the substance.