A 208 g sample of sodium-24 decays to 13.0 g of sodium-24 within 60.0 h. What is the half-life of this radioactive isotope?
To determine the half-life of Sodium-24, we need to analyze the given information.
We know that the initial mass of the sample is 208 g and that it decays to 13.0 g within a time of 60.0 hours.
The half-life of a radioactive isotope is the time it takes for half of the initial quantity to decay. In this case, we need to find the time it takes for 208 g to decay to 104 g (half of the initial mass).
The decay process can be modeled using the equation:
M = M₀ * (1/2)^(t / T½)
Where:
M = final mass (13.0 g in this case)
M₀ = initial mass (208 g)
t = time
T½ = half-life
Substituting the known values into the equation:
13.0 g = 208 g * (1/2)^(60.0 h / T½)
Now we can solve for the half-life, T½:
Dividing both sides by 208 g:
(1/2)^(60.0 h / T½) = 13.0 g / 208 g
Taking the logarithm of both sides:
log[(1/2)^(60.0 h / T½)] = log(13.0 g / 208 g)
Using the property of logarithms, we can rewrite the left side of the equation:
(60.0 h / T½) * log(1/2) = log(13.0 g / 208 g)
Now we can isolate T½ by dividing both sides by the logarithm of 1/2:
60.0 h / [log(1/2)] = log(13.0 g / 208 g)
Finally, solve for T½ by dividing 60.0 hours by the logarithm of 1/2 and substitute the values:
T½ = 60.0 h / log₂(1/2)
Using a calculator, we can calculate T½:
T½ ≈ 25.3 hours
Therefore, the half-life of Sodium-24 is approximately 25.3 hours.