Radium-221 has a half-life of 30 seconds. How long will it take for 94% of a sample to decay?

Feb 13, 2017

$t = 121.749 s$

Explanation:

Now, the equation for radiation of a radioactive substance is given by the equation
N=N_oe^{-lambdat
where $N$ implies the weight of the original radioactive mass remaining, ${N}_{o}$ being the original weight of the radioactive mass before the start of this experiment, $l a m \mathrm{da}$ being the degradation constant, and $t$ being time taken.

Now, apparently for the entire radioactive mass of Radium-221 taken to degrade such that only half of the original mass remains, 30 seconds pass by.
So, $t = 30 s$, $N = \left(1 - \frac{1}{2}\right) {N}_{o} = {N}_{o} / 2$, substituting them into the equation gives us
${\cancel{N}}_{o} / 2 = {\cancel{N}}_{o} {e}^{- 30 k}$

Applying ln (log base e) to both sides of the equation, I get
$- \ln 2 = - 30 \lambda$
So that means $\lambda = \ln \frac{2}{30}$ (further simplifying can be a hassle, so I'll keep it like this)

So the equation in total is $N = {N}_{o} {e}^{- \ln \frac{2}{30} t}$

Now, we're trying to find out how many seconds pass by until 94% of the total mass originally taken is degraded into something else. So, $N = \left(1 - \frac{94}{100}\right) {N}_{o} = \frac{6}{100} {N}_{o}$
Substituting that into the equation leads us to
$\frac{6}{100} {\cancel{N}}_{o} = {\cancel{N}}_{o} {e}^{- \ln \frac{2}{30} t}$
3/50=e^{-ln2/30t
Apply ln on both sides give us
$\ln \left(\frac{3}{50}\right) = - \ln \frac{2}{30} t$
You better get your calculator out here, it's turning out to be a tad windy.
$- 2.813 = - \frac{0.693}{30} t$

Now, the rest is up to you, solve this and save the human race from nuclear annihilation!