Question: Question 6 of 12, Step 2 of 3 The half-life of gold-195m is approximately 30.5 seconds. Step 2 of…

Question 6 of 12, Step 2 of 3

The half-life of gold-195m is approximately 30.5 seconds.

Step 2 of 3: How much of a 12 gram sample of gold-195m would remain after 41 seconds? Round to three decimal places.

Answer

How to enter your answer (opens in new window)

grams

Solution

The problem involves exponential decay, specifically radioactive decay, where the half-life of a substance is given. The formula to calculate the remaining quantity of a substance after a certain time is: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \] where: - \( N(t) \) is the remaining quantity after time \( t \), - \( N_0 \) is the initial quantity, - \( t \) is the elapsed time, - \( t_{1/2} \) is the half-life of the substance. Given: - \( N_0 = 12 \) grams, - \( t = 41 \) seconds, - \( t_{1/2} = 30.5 \) seconds. Substitute the known values into the formula: \[ N(41) = 12 \left(\frac{1}{2}\right)^{\frac{41}{30.5}} \] Calculate \( \frac{41}{30.5} \): \[ \frac{41}{30.5} \approx 1.3443 \] Calculate \( \left(\frac{1}{2}\right)^{1.3443} \): \[ \left(\frac{1}{2}\right)^{1.3443} \approx 0.4066 \] Now calculate \( N(41) \): \[ N(41) = 12 \times 0.4066 \approx 4.8792 \] Rounded to three decimal places, the amount of gold-195m remaining after 41 seconds is approximately 4.879 grams.