Question

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 socratic.org

Answer 1

Answer:

(i) Approximately 3 half lifes

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

Step-by-step explanation:

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

So, the number of half life in 11.46 days = $\frac{11.46}{3.8}$ ≈ 3

(ii) Since, the half life formula is,

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

Where,

$N_0$ = initial quantity,

t = number of periods

$t_{\frac{1}{2}}$ = half life of the quantity,

Given,

N = $5.2\times 10^{-8}\text{ g}$

t = 11.46 days,

$t_{\frac{1}{2}} = 3.8\text{ days}$

$\implies 5.2\times 10^{-8}=N_0 (\frac{1}{2})^\frac{11.46}{3.8}$

$\implies N_0=2^{\frac{11.46}{3.8}}\times 5.2\times 10^{-8}\approx 4.21\times 10^{-7}\text{ g}$