Answer: 33 minutes
Step-by-step explanation:
This question is considering the half-life of potassium-44, a radioactive substance, and wants to know how much time it takes for half of the initial amount of potassium-44 to decay. We can use the half-life formula and basic algebra to solve for half-life (T) based on the given information that it takes 100.0 minutes (t) for an initial amount of 20.00 g of potassium-44 at time 0 (N(0)) to decay to 2.50 g (N(t)):
N(t) = N(0) * (1/2)^(t/T), where
- N(t) = the amount of potassium-44 after 100.0 minutes of decay, 2.50 g
- N(0) = the initial amount of potassium-44 at 0 minutes, 20.00 g
- t = the time it took to go from N(0) to N(t), 100.0 minutes
- T = the half-life of potassium-44, which is what we are trying to solve
We have all the information needed, but we need to rearrange the above formula to solve for T, the half-life:
- N(t) = N(0) * (1/2)^(t/T)
- ( N(t) / N(0) ) = (1/2)^(t/T)
- You can then take the natural log (ln) of both sides to isolate the (t/T) term using the Power Rule: ln( N(t) / N(0) ) = (t/T) * ln(1/2)
- Solving for T:
T = t * ln(1/2) / ln( N(t) / N(0) )
Solving for T with the given information, we get:
T = [(100.0 mins) * ln(1/2) ] / [ ln( 2.50 g / 20.00 g) ] = 33 minutes
Logically, we can check this answer. If we start with 20.00 g, after 33 minutes, we have 10.00 g; for another 33 minutes (total time=66 minutes), we will have 5.00g, and with an additional 33 minutes (total time=99 minutes, which is close to the given 100.0 minutes), we will have 2.50 g.