Question

How much time is required for a \(6.25-\mathrm{mg}\) sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75 \mathrm{mg}\) if it has a half-life of \(27.8\) days?

Short answer

It will take approximately \(83.4\) days for the \(6.25 \, \text{mg}\) sample of Chromium-51 to decay to \(0.75 \, \text{mg}\).

Step-by-step solution

Identify given information

We have the following information given in the exercise: - Initial amount, \(N_0 = 6.25 \, \text{mg}\) - Final amount, \(N(t) = 0.75 \, \text{mg}\) - Half-life, \(T = 27.8 \, \text{days}\)

Write the decay formula

Using the exponential decay formula, we have: \(0.75 = 6.25 \cdot (1/2)^{t/27.8}\)

Solve for t

Now our goal is to find the value of t (the time required for the decay) by solving the above equation. First, divide both sides by 6.25: \(\frac{0.75}{6.25} = (1/2)^{t/27.8}\) Next, simplify the left side: \(\frac{1}{8} = (1/2)^{t/27.8}\) Now, we can take the base-2 logarithm of both sides: \( \log_2 \frac{1}{8} = \log_2 (1/2)^{t/27.8} \) Using the logarithm property, we get: \( -3 = \frac{t}{27.8}\) Finally, multiply both sides by 27.8 to solve for t: \(t = -3 \cdot 27.8\) \(t \approx -83.4 \, \text{days}\)

Interpret the result

The negative sign in the result indicates an error in our calculations (the time cannot be negative). This error arises from missing a step while applying the logarithm property. Let's fix the mistake and recalculate t in the correct way.

(fixed): Solve for t

We have: \(\frac{1}{8} = (1/2)^{t/27.8}\) Taking the base-2 logarithm of both sides, we get: \( \log_2 \frac{1}{8} = \log_2 (1/2)^{t/27.8} \) Using the logarithm property, we have: \( -3 = \frac{t}{27.8} \cdot \log_2(1/2)\) In this step, we should apply the property of logarithms, which states that \(\log_b(a^c) = c \cdot \log_b(a)\), but we missed it previously. Since \(\log_2(1/2) = -1\), our equation now becomes: \(-3 = -\frac{t}{27.8}\) Now, solve for t: \(t = 3 \cdot 27.8\) \(t \approx 83.4 \, \text{days}\) So, it will take approximately 83.4 days for the 6.25 mg sample of Chromium-51 to decay to 0.75 mg.