The half life of iodine - 131 is 8 days. How long will it take for 7/8 of the sample to decay?
That means 1/8 is left.
1/8=(1/2)^n
looks like n=3days
To find out how long it will take for 7/8 of the sample to decay, we can use the concept of the half-life.
The half-life of iodine-131 is given as 8 days, which means that after every 8 days, half of the sample will decay. So, we need to determine how many half-lives are required for 7/8 (or 87.5%) of the sample to decay.
To do this, we can use the following formula:
N(t) = N₀ * (1/2)^(t/h)
Where:
- N(t) is the remaining amount of sample after time t
- N₀ is the initial amount of the sample
- t is the time elapsed
- h is the half-life of the substance
Let's assume the initial amount of the sample is 100 (just for calculation purposes).
N(t) = 100 * (1/2)^(t/8)
Since we want 7/8 (or 87.5%) of the sample to decay, we can set N(t) equal to 87.5 and solve for t.
87.5 = 100 * (1/2)^(t/8)
Now, we can solve this equation for t. Taking the logarithm of both sides (base 1/2) will help us isolate the exponent t/8.
log₀.₅ (87.5) = log₀.₅ (100 * (1/2)^(t/8))
To simplify the equation, we can use the logarithmic property that states log(b * c) = log(b) + log(c).
log₀.₅ (87.5) = log₀.₅ (100) + log₀.₅ ((1/2)^(t/8))
Since log₀.₅ (100) = 0 (any number raised to 0 is 1), we have:
log₀.₅ (87.5) = log₀.₅ ((1/2)^(t/8))
Now, we can rewrite the equation as an exponent equation:
(1/2)^(t/8) = 87.5
To find t, we need to isolate the t/8 exponent by taking the logarithm of both sides (base 1/2).
log₀.₅ [(1/2)^(t/8)] = log₀.₅ (87.5)
Using the exponent property of logarithms, we can move the t/8 exponent down:
(t/8) * log₀.₅ (1/2) = log₀.₅ (87.5)
Knowing that log(x) / log(y) = log(y, x), we can rewrite this equation as:
(t/8) = log₀.₅ (87.5) / log₀.₅ (1/2)
Now, we can evaluate the right side of the equation using a calculator:
(t/8) ≈ (-0.14612) / (-0.30103)
Dividing gives:
(t/8) ≈ 0.485
Multiplying both sides by 8:
t ≈ 3.88
Therefore, it will take approximately 3.88 (or 4) days for 7/8 of the sample to decay.