if the amount of radioactive iodine-123 in a sample decreases from .400g to .100g in 26.2 hours, what is the half-life of iodine-123
ln(No/N)=kt
Substitute No = 0.400
N = 0.1
k = unknown.
t = 26.2 hrs.
Then k = 0.693/t1/2
You know k, solve for t1/2
To determine the half-life of iodine-123, we need to utilize the exponential decay formula. The formula for exponential decay is:
A = A₀ * (1/2)^(t / t₁/₂)
Where:
A is the final amount of substance.
A₀ is the initial amount of substance.
t is the elapsed time.
t₁/₂ is the half-life of the substance.
In this case, we know that:
A₀ = 0.400g (initial amount)
A = 0.100g (final amount)
t = 26.2 hours.
We can plug these values into the formula and solve for t₁/₂:
0.100g = 0.400g * (1/2)^(26.2 / t₁/₂)
Now let's isolate t₁/₂:
(1/2)^(26.2 / t₁/₂) = 0.100g / 0.400g
(1/2)^(26.2 / t₁/₂) = 0.25
To remove the exponent, we take the logarithm of both sides. Let's use the natural logarithm (ln) to solve for t₁/₂:
ln((1/2)^(26.2 / t₁/₂)) = ln(0.25)
Using the logarithmic properties, the exponent can be brought down:
(26.2 / t₁/₂) * ln(1/2) = ln(0.25)
Now, solve for t₁/₂ by isolating it:
t₁/₂ = (26.2 * ln(1/2)) / ln(0.25)
Using a calculator, evaluate the right side of the equation to find the value of t₁/₂.
Please note that ln(1/2) is approximately -0.6931, and ln(0.25) is approximately -1.3863.
Once you perform the calculations, you will obtain the value of t₁/₂, which represents the half-life of iodine-123 in this scenario.