How do I find an equation for the amount of Thorium$-234$ at any time $t$?

Question

The isotope of Thorium-$234$ has a half-life of $24.5$ days. Find an equation which gives the amount of Thorium-$234$ at any time $t$ for a sample with initial size $2$ kg. At what time does the sample weight $0.75$ kg?

$$\ y = Ce^{-kt} \ $$ $$\ y = 2e^{-k(24.5)} \ $$ $$\ \ln{y} = \ln{\Big(6e^{-k(24.5)}\Big)} \ $$ $$\ \ln{y} = -6k(24.5) \ $$

I know to solve for $k$. However, if $y$ is unknown, then I can only get so far. Must I use $0.75$ kg as $y$ to find $k$? The above approach is incorrect, I am almost positive.

Is the following the correct approach:

$$\ T_{1/2} = \frac{\ln{2}}{k} \ $$ $$\ k = \frac{\ln2}{T_{1/2}} = \frac{\ln{2}}{24.5} \approx 0.028 \ $$

Giving me

$$\ y = 2e^{-0.028t} \ $$ $$\ 0.75 = 2e^{-0.028t} \ $$ $$\ t = \frac{\ln(0.375)}{-0.028} \approx 35.0 \ $$

I want to clear myself of doubt of whether this is correct.

You are all appreciated!

Thanks,

Michael

Answer 1

Yes, your second approach is correct. In the first approach, I'm not sure exactly how you got the $6$ to be the constant, and you would not be using half-life but rather $e$ as the exponential base. Not quite what you want. Also, when taking the natural logarithm of both sides, the $6$ disappeared. But yes, the approach you took the second time is correct, but I would suggest using the exact value instead of $-0.28$ and use $\frac{\ln(2)}{24.5}$ to avoid round-off.