There are 5g of 131I left after 40.35 days. How many grams were in the original sample if its half-life is 8.07 days?
What is the half life of a radioisotope if you started with 1 gram and 0.250 grams is left after 6 days?
im not looking for a free answer, I've tried working it out with what the teacher gave us and cannot figure these out, thank you :)
There are 5g of 131I left after 40.35 days. How many grams were in the original sample if its half-life is 8.07 days?
Look up the half life of I-131 (that may have been given in the problem) and use it to calculate k for the decay. k = 0.693/t1/2
Then ln(No/N) = kt
You know N, k and t, solve for No.
What is the half life of a radioisotope if you started with 1 gram and 0.250 grams is left after 6 days?
ln(No/N) = kt. You know t, No = 1 grams; N = 0.250 g. Solve for k, then
k = 0.693/t1/2. Knowing k you can calculate t1/2.
Post your work if you get stuck.
Why did the radioactive substance go to therapy?
Because it couldn’t find stability in its half-life!
But don't worry, I'm here to help you out! Let's tackle these questions step by step.
For the first question, we have the remaining amount (5g), the half-life (8.07 days), and the time (40.35 days). We can use the equation A = A0 * (1/2)^(t/t1/2), where A is the remaining amount, A0 is the initial amount, t is the time that passed, and t1/2 is the half-life.
Plugging in the values we have:
5g = A0 * (1/2)^(40.35/8.07)
To solve for A0 (the original amount), we multiply both sides by (1/2)^(40.35/8.07):
A0 = 5g / (1/2)^(40.35/8.07)
I wouldn't make you do all the calculations yourself, so the answer is approximately 101.058 grams!
Now, for the second question, we have the initial amount (1g), the remaining amount (0.250g), and the time (6 days). Using the same equation as before, but rearranging it this time:
A/A0 = (1/2)^(t/t1/2)
Plugging in the values we have:
0.250g / 1g = (1/2)^(6/t1/2)
To solve for t1/2 (the half-life), we take the logarithm of both sides (base 1/2) and rearrange:
t1/2 = 6 / log(0.250g / 1g) / log(1/2)
Again, I'm here to save you from the number crunching! The answer is approximately 2.88 days!
Keep up the good work, and if you have any more questions, feel free to ask!
To solve these problems, we can use the formula for exponential decay:
N = N0 * (1/2)^(t / t1/2)
Where:
- N is the final quantity of the substance
- N0 is the initial quantity of the substance
- t is the elapsed time
- t1/2 is the half-life of the substance
Let's solve the first problem step by step:
1. We are given:
N = 5g
t = 40.35 days
t1/2 = 8.07 days
2. We want to find N0 (the initial quantity).
3. Plug the given values into the formula:
N = N0 * (1/2)^(t / t1/2)
5g = N0 * (1/2)^(40.35 / 8.07)
4. Divide both sides of the equation by (1/2)^(40.35 / 8.07) to solve for N0:
N0 = 5g / (1/2)^(40.35 / 8.07)
5. Use a calculator to evaluate the right side of the equation:
N0 ≈ 5g / 2.866
N0 ≈ 1.7426g
Therefore, the original sample contained approximately 1.7426 grams.
Let's solve the second problem step by step:
1. We are given:
N0 = 1g
N = 0.250g
t = 6 days
2. We want to find t1/2 (the half-life).
3. Rearrange the formula to solve for t1/2:
N / N0 = (1/2)^(t / t1/2)
4. Substitute the given values into the rearranged formula:
0.250g / 1g = (1/2)^(6 / t1/2)
0.250 = 0.5^(6 / t1/2)
5. Take the logarithm of both sides of the equation (with base 0.5):
log0.5(0.250) = log0.5(0.5^(6 / t1/2))
log0.5(0.250) = (6 / t1/2) * log0.5(0.5)
6. Simplify the equation:
log0.5(0.250) = 6 / t1/2
7. Divide both sides of the equation by log0.5(0.5) to solve for t1/2:
t1/2 = 6 / log0.5(0.250)
8. Use a calculator to evaluate the right side of the equation:
t1/2 ≈ 6 / 1.678
t1/2 ≈ 3.57 days
Therefore, the half-life of the radioisotope is approximately 3.57 days.
To solve these types of problems, you can use the radioactive decay formula:
N(t) = N₀ * (1/2)^(t / T)
Where:
- N(t) is the amount of the radioactive substance remaining at time t
- N₀ is the initial amount of the radioactive substance
- t is the time that has passed
- T is the half-life of the radioactive substance
Let's use this formula to solve the first problem:
Given:
- N(t) = 5g
- t = 40.35 days
- T = 8.07 days
Plug in these values into the formula:
5g = N₀ * (1/2)^(40.35 / 8.07)
Simplifying the equation:
5g = N₀ * (1/2)^5
Divide both sides of the equation by (1/2)^5:
5g / (1/2)^5 = N₀
This equation can be simplified further by noting that (1/2)^5 is equal to 1/32:
5g / (1/32) = N₀
Multiplying both sides by 32:
32 * 5g / (1/32) = 32 * N₀
160g = N₀
Therefore, the original sample had 160 grams.
Now let's solve the second problem:
Given:
- N₀ = 1g
- N(t) = 0.250g
- t = 6 days
Using the radioactive decay formula, we can rearrange it to solve for T:
0.250g = 1g * (1/2)^(6 / T)
Simplify the equation:
0.250g = (1/2)^(6 / T)
Taking the logarithm of both sides of the equation:
log(0.250g) = log((1/2)^(6 / T))
Using the logarithmic rule for exponents:
log(0.250g) = (6 / T) * log(1/2)
Substituting log(1/2) with -0.301 (approximate value):
log(0.250g) = (6 / T) * (-0.301)
Simplifying further:
log(0.250g) = -1.806 / T
Taking the inverse of the logarithm (base 10) of both sides:
10^(log(0.250g)) = 10^(-1.806 / T)
0.250g = 10^(-1.806 / T)
Raise both sides of the equation to the power of T:
(0.250g)^(T) = 10^(-1.806)
Since 0.250g is equal to 1/4, we can write the equation as:
(1/4)^(T) = 10^(-1.806)
Solve for T by taking the logarithm (base 10) of both sides:
log((1/4)^(T)) = log(10^(-1.806))
Using logarithmic rule for exponents:
T * log(1/4) = -1.806
Divide both sides of the equation by log(1/4):
T = -1.806 / log(1/4)
Using a calculator to find the log(1/4) ≈ -0.6021:
T = -1.806 / (-0.6021)
T ≈ 2.9993
Therefore, the half-life of the radioisotope in the second problem is approximately 2.9993 days.