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.

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