What is the halflife of sodium 25 if 1.00 gram of a 16.00 gram sample of sodium 25 remains unchanged after 237 seconds?

Answer 1

59.25 seconds

After 1 half-life: 16*(1/2) = 8 g remains

After 2 half-lives: 8*(1/2) = 4g remains

After 3 half-lvies: 4*(1/2) = 2g remains

After 4 half-lives: 2*(1/2) = 1g remains

So after 4 half-lves you have 1 gram of Na25 left. This is also the amount remaining after 237 seconds. Since 4 half-lives have elapsed over 237 seconds, you can divide 237 seconds by 4 to find the half-life of Na25 is 59.25 seconds.

You can also figure it out using the rate of decay formula:

At = Ao*e-kt

where Ao is the initial amount, At is the amount left after some time t. k is the decay constant which is k = ln 2 / t1/2 where t1/2 is the half-life.

In this case Ao = 16g, At = 1g, t=237 sec

Substitute in the formula to solve for k, then take the answer for k and use it in the other formula to solve for the half-life (t1/2):

1 = 16*e-237k

ln (1/16) = ln (e-237k)

-2.7726 = -237k

k = -2.7726/-237 = 0.011699 sec-1

t1/2 = ln 2/k = 0.693147/0.011699 sec-1

t1/2 = 59.25 sec