What is the rate of decay from 1.00 mol of radioactive nuclides having the following half-lives: \(12,000\) years? 12 hours? 12 seconds?

We will convert the given half-lives to decay constants using the formula: \( k = \frac{ln(2)}{T} \), where T is the half-life and k is the decay constant. 1. For 12,000 years: \( k = \frac{ln(2)}{12000} \) 2. For 12 hours: Convert 12 hours to years: \( 12 \frac{hours}{1\ day} \times \frac{1\ day}{24\ hours} \times \frac{1\ year}{365.25\ days} = \frac{1}{730.5}\ years \) \( k = \frac{ln(2)}{\frac{1}{730.5}} \) 3. For 12 seconds: Convert 12 seconds to years: \( 12 \frac{seconds}{1\ minute} \times \frac{1\ minute}{60\ seconds} \times \frac{1\ hour}{60\ minutes} \times \frac{1\ day}{24\ hours} \times \frac{1\ year}{365.25\ days} = \frac{1}{31,622,400}\ years \) \( k = \frac{ln(2)}{\frac{1}{31,622,400}} \)

We will now calculate the decay rates for each half-life by multiplying the decay constant (k) by the initial number of moles (1 mol). 1. For 12,000 years: Decay Rate = \( k \times 1\ mol = \frac{ln(2)}{12000} \times 1\ mol \) 2. For 12 hours: Decay Rate = \( k \times 1\ mol = \frac{ln(2)}{\frac{1}{730.5}} \times 1\ mol \) 3. For 12 seconds: Decay Rate = \( k \times 1\ mol = \frac{ln(2)}{\frac{1}{31,622,400}} \times 1\ mol \)

Finally, we will simplify the expressions for the decay rates. 1. For 12,000 years: Decay Rate = \( \frac{ln(2)}{12000}\ mol\ year^{-1} \) 2. For 12 hours: Decay Rate = \( 730.5 \times ln(2)\ mol\ year^{-1} \) 3. For 12 seconds: Decay Rate = \( 31,622,400 \times ln(2)\ mol\ year^{-1} \) So, the decay rates for the given half-lives are: 1. For 12,000 years: \( \frac{ln(2)}{12000}\ mol\ year^{-1} \) 2. For 12 hours: \( 730.5 \times ln(2)\ mol\ year^{-1} \) 3. For 12 seconds: \( 31,622,400 \times ln(2)\ mol\ year^{-1} \)