Exponential Decay: Solving with Radioactive Half-Lives

[chroot]

From a member called "Banana":

Now I'm trying to do one with radioactive decay. Do you think I would use those same formulas? The only confusing thing is that it's not presented the same. It says that a material decays so that it is 99 percent gone in 6.65 half-lives (so would you double that?). After how many half-lives is the material 99.9 percent gone? I tried to piece together some ideas I found in other textbooks, but I don't think it makes sense. Here's what I tried:

- ln (N/No)/ln2
- ln (.001/.01)/.693 = 3.3 (I don't think this is right because I didn't even find anywhere to insert the 6.65)

Is that wrong? Thanks for your help, I REALLY appreciate it. Banana

 

[chroot]

The key equation is the same. In this case, after one half-life has elapsed, the population of remaining radioactive atoms is one-half what it was at time t=0.

Thus

$$0.5 N_0 = N_0 e^{k t_{hl}}$$

Where $t_{hl}$ is the half-life.

If one-half the population remains after one half-life, you get this equation:

$$0.5 = e^{k \cdot 1}$$

Thus, k always equals $\ln{0.5}$, or about -0.69, in problems dealing with half-life.

Now, using this k, we can solve for the situation where 99.9% of the original population is gone (therefore 0.1% remains):

$$0.001 N_0 = N_0 e^{\ln{0.5} t}$$

Solve for t. Let me know what you get.

- Warren

 

[himanshu121]

$$\frac{a}{a-x}=2^{\frac{t}{\tau}}$$

tau : half life

And a is initial amt and a-x is the remaining
from the data

$$log(1000) =log2*\frac{t}{\tau}$$

So u get t={log1000/log2}*tau =9.96 $$\tau$$