The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decompose after 5.55 days if the sample initially weighs 100.0 grams?

Question

Mateo Aguilar · Accepted Answer

After 5.55 days, 63.4 g of radon will have decayed.

A first-order process's half-life can be calculated using the formula #color(blue)(bar(ul(|color(white)(a/a) t_½ = (ln2)/kcolor(white)(a/a)|)))" "# ∴ #k = (ln2)/t_½ = (ln2)/"3.823 days" = "0.1813 day"^"-1"# A first-order process's integrated formula is #color(blue)(bar(ul(|color(white)(a/a) ln"A"_0 - ln"A" = ktcolor(white)(a/a)|)))" "# #ln100 - ln"A" = kt = 0.1813 color(red)(cancel(color(black)("day"^"-1"))) × 5.55 color(red)(cancel(color(black)("day"))) = 1.006# #ln"A" = ln100 - 1.006 = 3.599# #"A" = e^3.599color(white)(l) "g" = "36.55 g"# If 36.55 g are still present, the mass that broke down is #"A"_0 - "A" = "100.0 g - 36.55 g" = "63.4 g"#

Peyton Adams · Answer

undefined To find the amount of radon decomposed after 5.55 days, use the first-order decay equation:

$$ A = A_0 \times e^{-\lambda t} $$

Where:
- $ A $ is the final amount of radon after $ t $ days
- $ A_0 $ is the initial amount of radon
- $ \lambda $ is the decay constant
- $ t $ is the time in days

Given:
- $ A_0 = 100.0 $ grams
- $ t = 5.55 $ days
- Half-life $ T_{1/2} = 3.823 $ days

First, calculate the decay constant $ \lambda $ using the half-life formula:

$$ T_{1/2} = \frac{\ln(2)}{\lambda} $$

Then, plug in the values to find $ \lambda $. Once you have $ \lambda $, use it in the decay equation to find $ A $.