If the amount of radioactive iodine-123 in a sample decreases from 0.8 g to 0.1 g in 36.2 hours, what is the half-life of iodine-123
Bot, Please answer this question
Find the interquartile range (IQR) of the data
98 132 295 68 176 89 287 198 94 68 240 94 369 94 155
Bot
To find the half-life of iodine-123, you can use the formula for radioactive decay:
N(t) = N₀ * (1/2)^(t / T₁/₂)
Where:
N(t) = the remaining amount of substance at time t
N₀ = the initial amount of substance
t = elapsed time
T₁/₂ = half-life of the substance
In this case, the initial amount of radioactive iodine-123 (N₀) is 0.8 g, and the remaining amount at a later time (N(t)) is 0.1 g. The elapsed time (t) is 36.2 hours.
We can rearrange the formula to solve for the half-life:
(1/2)^(t / T₁/₂) = N(t) / N₀
Taking the logarithm of both sides will isolate the exponent:
log[(1/2)^(t / T₁/₂)] = log(N(t) / N₀)
Using the logarithmic property, we can bring the exponent down:
(t / T₁/₂) * log(1/2) = log(N(t) / N₀)
Now, we can solve for the half-life (T₁/₂):
T₁/₂ = (t * log(1/2)) / log(N(t) / N₀)
Substituting the given values:
T₁/₂ = (36.2 hours * log(1/2)) / log(0.1 g / 0.8 g)
Calculating the logarithms and performing the division:
T₁/₂ = (36.2 hours * -0.693) / -0.699
T₁/₂ ≈ 36.2 hours * 0.993
T₁/₂ ≈ 35.94 hours
Therefore, the half-life of iodine-123 is approximately 35.94 hours.