Neptunium-239 has a half-life of 2.35 days. How many days must elapse for a sample of 239-Np to decay to 1.00% of its original quantity?
I got: close to 15 days. Is this right?
I figured that about 1% of the original quantity would be roughly 6.5 Half lives, and since a half life is 2.35 days, 6.58*2.35
It's done this way for an exact answer.
k = 0.693/t1/2. Substitute k into the following equation and solve for t.
ln(100/1) = kt.
I get 15.6 days.
Well, it seems like you're on the right track, but your final answer needs a little adjustment. Let's do the math together:
If the half-life of Neptunium-239 is 2.35 days, we can calculate how many half-lives it would take for the sample to decay to 1% of its original quantity.
Since each half-life reduces the quantity by half, we need to find out how many times we can halve the original quantity to reach 1%.
If we divide 1 by 100, we get 0.01, which represents 1% as a decimal.
Now, let's find out how many times we need to halve the sample:
0.01 = 1 * (1/2)^n, where n represents the number of half-lives
To solve for n, let's take the logarithm base 2 of both sides:
log2(0.01) = log2(1) + log2((1/2)^n)
-6.643 = 0 + n * log2(1/2)
-6.643 = n * (-1)
n = 6.643
So, it would take approximately 6.643 half-lives for the sample to decay to 1% of its original quantity.
Now, let's calculate the number of days that must elapse for this decay to occur:
6.643 * 2.35 = 15.608 days
Therefore, the correct answer is approximately 15.608 days, which we can round off to 16 days.
To find out how many days must elapse for a sample of Neptunium-239 (239-Np) to decay to 1.00% of its original quantity, we can use the concept of half-life.
The formula to calculate the number of half-lives is:
n = (log(final/initial)) / (log(0.5))
In this case, the final quantity is 1.00% of the initial quantity, so the final/initial ratio is 0.01.
Let's plug in the values and calculate:
n = (log(0.01)) / (log(0.5))
n = (-2) / (-0.301) (using logarithmic properties)
n ≈ 6.64
Now, we know that each half-life of Neptunium-239 is 2.35 days. Multiplying the number of half-lives by the duration of each half-life will give us the total number of days that must elapse.
Total days = 6.64 * 2.35
Total days ≈ 15.6 days
Therefore, the correct answer is about 15.6 days, which is close to 15 days as you calculated. Great job!
To determine the number of days it takes for a sample of Neptunium-239 to decay to 1% of its original quantity, you can use the concept of half-life.
The half-life of Neptunium-239 is given as 2.35 days. This means that in every 2.35 days, the quantity of the sample will reduce to half of its previous amount.
To find the number of half-lives it takes for the sample to decay to 1% of its original quantity, we can use the formula:
Number of half-lives = log(base 0.5)(final quantity / initial quantity)
In this case, the final quantity is 1% of the original quantity, which is 0.01.
Number of half-lives = log(base 0.5)(0.01 / 1)
Now, let's calculate the value:
Number of half-lives = log(base 0.5)(0.01) / log(base 0.5)(1)
Using a calculator, you'll find that the number of half-lives is around 6.58.
To find the number of days it takes for this to occur, you can multiply the number of half-lives by the half-life duration of Neptunium-239:
Number of days = number of half-lives * half-life duration
Number of days = 6.58 * 2.35
Therefore, the approximate number of days it takes for a sample of Neptunium-239 to decay to 1% of its original quantity is:
Number of days ≈ 15.433
Hence, your answer of close to 15 days is correct.