Question

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many half lived have passed?

Answer 1

Approximately 3 half-lives of radon-222 have passed for the sample to decay from 150mg to 18.75mg over the course of 11.4 days, given its half-life of 3.823 days.

Step-by-step explanation:

The student's question is about determining how many half-lives have passed during the decay of radon-222 (Rn-222) from an initial mass of 150mg to 18.75mg after 11.4 days. To calculate the number of half-lives, we use the half-life formula:

N_t = N_0 (1/2)^(t/T)

• N_t is the remaining quantity after time t.
• N_0 is the initial quantity.
• t is the total time elapsed.
• T is the half-life of the substance.

In this case, the half-life (T) of radon-222 is 3.823 days. We're given the initial mass (N_0) of 150mg, the remaining mass (N_t) of 18.75mg, and the elapsed time (t) of 11.4 days. After substituting the values into the above formula and solving for the number of half-lives, we find:

18.75 = 150(1/2)^(11.4/3.823)

Simplifying, we can calculate the number of half-lives (n):

n = 11.4 / 3.823

n ≈ 2.982

Therefore, approximately 3 half-lives have passed.

Answer 2

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

Step-by-step explanation:

Given, the initial value of the sample,
`A_0` = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

`(A)/(A_(0))=\left((1)/(2)\right)^(N)`

where N is the no. of half life

Substituting the values,

`(18.75)/(150)=\left((1)/(2)\right)^(N)`

`\left((1)/(2)\right)^(3)=\left((1)/(2)\right)^(N)`

On equating, we get, N = 3

Therefore, 3 half-lives have passed.