How many half-lives of radon-222 have passed in 11.46 days? If 5.2 × 10−8 g of radon-222 remain in a sealed box after 11.46 days, how much was present in the box initially socra…

Question

asked Sep 18, 2020 118k views

1 Answer

Molicule · Answer 1 · 2020-09-23T22:50:27+0000

Answer:

(i) Approximately 3 half lifes

(ii)
$4.21* 10^(-7)\text{ g}$

Explanation:

(i) ∵ The half life of Radon-222 is approximately 3.8 days,

So, the number of half life in 11.46 days =
(11.46)/(3.8) ≈ 3

(ii) Since, the half life formula is,

$N=N_0 ((1)/(2))^{\frac{t}{t_{(1)/(2)}}}$

Where,

N_0 = initial quantity,

t = number of periods

$t_{(1)/(2)}$ = half life of the quantity,

Given,

N =
$5.2* 10^(-8)\text{ g}$

t = 11.46 days,

$t_{(1)/(2)} = 3.8\text{ days}$

$\implies 5.2* 10^(-8)=N_0 ((1)/(2))^(11.46)/(3.8)$

$\implies N_0=2^{(11.46)/(3.8)}* 5.2* 10^(-8)\approx 4.21* 10^(-7)\text{ g}$