Question: The half-life of gold-195m is approximately 30.5 seconds. Step 3 of 3: How much of a 12 gram…

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

Step 3 of 3: How much of a 12 gram sample of gold-195m would remain after 79 seconds? Round to three decimal places.

Answer: __ grams

Solution

The problem involves calculating how much of a 12 gram sample of gold-195m remains after 79 seconds, given its half-life of 30.5 seconds. We’ll use the half-life decay formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \] where: - \( N(t) \) is the remaining quantity of the substance after time \( t \), - \( N_0 \) is the initial quantity of the substance, - \( T \) is the half-life of the substance, - \( t \) is the time elapsed. Start by plugging in the given values: \( N_0 = 12 \) grams, \( T = 30.5 \) seconds, \( t = 79 \) seconds. First, calculate the exponent: \[ \frac{t}{T} = \frac{79}{30.5} \] \[ \frac{79}{30.5} \approx 2.590 \] Now, calculate the remaining quantity: \[ N(t) = 12 \left(\frac{1}{2}\right)^{2.590} \] \[ N(t) \approx 12 \times 0.169 \] \[ N(t) \approx 2.028 \] Therefore, approximately 2.028 grams of gold-195m would remain after 79 seconds.