Question

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 95% of the lead to decay?

Answer 1

Answer:

$N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}$

Where $t_{1/2}= 3.3 hr$ represent the half life and the intial amount would be $N_o = 1$

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

$0.05 = 1 (0.5)^{t/3.3}$

If we apply natural log on both sides we got:

$ln(0.05) = \frac{t}{3.3} ln (0.5)$

And solving for t we got:

$t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26$

So then would takes about 14.26 hours in order to have  95% of the lead to decay

Step-by-step explanation:

For this case we can define the variable of interest amount of Pb209 and for the half life would be given:

$N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}$

Where $t_{1/2}= 3.3 hr$ represent the half life and the intial amount would be $N_o = 1$

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

$0.05 = 1 (0.5)^{t/3.3}$

If we apply natural log on both sides we got:

$ln(0.05) = \frac{t}{3.3} ln (0.5)$

And solving for t we got:

$t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26$

So then would takes about 14.26 hours in order to have  95% of the lead to decay