In a tracer method the potassium isotope 42K is used for labelling. The half life of 42K is 12.5 hours. If No is original number of atoms, how many hours will it take until only (1/1024) No Atoms remain?

The answer is 125 hours but I do not know how to get this. Please help!

Goodness.

change the prior problem to

1/1024=(1/2)^(t/12.5)

2 to the what = 1024 ?


(If you do not know instantly do this
2^x = 1024 )

x log 2 = log 1024
x = 10
so
(1/2)^10 = 1/1024
t/12.5 = 10
t = 125

To determine the number of hours it will take until only 1/1024th of the original number of atoms remains, we can use the concept of half-life.

In this case, we know that the half-life of the potassium isotope 42K is 12.5 hours. This means that every 12.5 hours, half of the 42K atoms will decay, and the remaining half will remain.

Let's break down the problem step by step:

1. Start with the initial number of atoms, which is No.
2. After 12.5 hours, the number of remaining atoms will be 1/2 of No, which is No/2.
3. After another 12.5 hours (totaling 25 hours), the number of remaining atoms will be 1/4 of No, which is No/2^2.
4. Continuing this pattern, after 37.5 hours (12.5 + 12.5 + 12.5), the number of remaining atoms will be 1/8 of No, which is No/2^3.
5. Generalizing, after t hours, the number of remaining atoms will be No/2^(t/12.5).
6. To find the time it takes until only 1/1024th of the original atoms remain, we need to solve the equation No/2^(t/12.5) = No/1024. Cross-multiplying, we get 2^(t/12.5) = 1024.
7. We can rewrite 1024 as 2^10, so the equation becomes 2^(t/12.5) = 2^10.
8. Since the bases are the same, we can equate the exponents: t/12.5 = 10.
9. Solving for t, we get t = 10 * 12.5 = 125 hours.

Therefore, it will take 125 hours until only 1/1024th of the original number of atoms remain.