Neptunium-239 has a half-life of 2.35 days. How many days must elapse for a sample of 239Np to decay to 0.100% of its original quantity
The half life equation is:
N(t)/N(0) = (1/2)^(t/2.35)
Solve for t when: N(t)/N(0) = 0.00100
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23.4 days
To determine the number of days required for a sample of Neptunium-239 (239Np) to decay to 0.100% of its original quantity, we need to use the concept of half-life.
The half-life of a radioactive substance is the amount of time it takes for half of the original quantity to decay. In this case, the half-life of 239Np is given as 2.35 days.
To solve this problem, we can use the following equation:
N(t) = N0 * (1/2)^(t / T)
Where:
N(t) = final quantity of the substance
N0 = initial quantity of the substance
t = time elapsed
T = half-life of the substance
We need to find the time (t) when the final quantity (N(t)) is equal to 0.100% (0.001) of the initial quantity (N0).
0.001 = N0 * (1/2)^(t / 2.35)
Now, let's solve for t:
(1/2)^(t / 2.35) = 0.001 / N0
Taking the logarithm on both sides:
log10((1/2)^(t / 2.35)) = log10(0.001 / N0)
Using the logarithmic identity, log(a^b) = b * log(a):
(t / 2.35) * log10(1/2) = log10(0.001 / N0)
(t / 2.35) * (-0.301) = log10(0.001 / N0)
Simplifying further:
t = (2.35 * log10(0.001 / N0)) / (-0.301)
Now, plug in the value for N0, which is 100% or 1, into the equation:
t = (2.35 * log10(0.001)) / (-0.301)
Using a calculator, we can find:
log10(0.001) ≈ -3
t = (2.35 * -3) / (-0.301)
t ≈ 23.42 days
Therefore, it would take approximately 23.42 days for a sample of 239Np to decay to 0.100% of its original quantity.