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?

1 Answer
Oct 16, 2016

Answer:

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

Explanation:

The formula for the half-life of a first-order process is

#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"#

The integrated formula for a first-order process 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 remain, the mass that decomposed is

#"A"_0 - "A" = "100.0 g - 36.55 g" = "63.4 g"#