Question

Radon-222 (222 86 Rn) is a radioactive gas that is often found in homes as a byproduct of natural radiation in the ground. It has a mass of 222.017578u and a half-life of 3.8235 days. You are conducting an experiment on the effects of Radon on bacteria. You begin the experiment at 8:00 am on Monday when you introduce 4.00 µg of Radon-222. Later in the week you are concerned that the activity is too low for your experiment. How many atoms of Radon remain at 2:00 pm on Thursday? What is the activity? After doing these calculations, you add another 4.00 µg of Radon to the experiment (at 2:00 pm on Thursday). What is the activity when you arrive at work at 8:00 am the following Monday?

Answer 1

Solution:

We know that, k, decay constant = 0.693/t_1/2

                                         k = 0.693 / 3.8235 = 0.1812 day^-1

From Monday 08.00 am to Thursday 2 pm it is 78 hrs or 78/24 days = 3.25 days

t = 3.25 days

k = 0.1812 day^-1

A₀ = 4 $\mu$ g

A_t = ?

2.303 *log0 At t k

kt log 40 2.303 At

             = 0.1812 x 3.25 / 2.303 = 0.2557

A₀ / A_t = Anti log (0.2557) = 1.80177

A_t = A₀ / 1.80177 = 4 / 1.8017 = 2.22 $\mu$ g

The activity on Thursday 2 pm is 2.22 x 100/4 = 55.5%

Number of atoms of Radan present is

2.22 106.023 1023 222

                                                 = 6.023 x 10¹⁵ atoms.

4 $\mu$ g added to this so that the total will be 6.22 $\mu$ g on thursday 2.00 pm.

From Thursday 2.00 pm to Monday 08.00 am it is 90 hrs = 3.75 days

quantity of radan remaining on monday 8.00 am will be ( following the same previous steps)

t = 3.75 days

k = 0.1812 day^-1

A₀ = 6.22 $\mu$ g

A_t = ?

2.303 *log0 At t k

kt log 40 2.303 At

     = 0.1812 x 3.75 / 2.303 = 0.295

A₀ / A_t = Anti log (0.295) = 1.972

A_t = A₀ / 1.972 = 6.22 / 1.972 = 3.154 $\mu$ g

The activity will be 3.154 * 100 /6.22 = 50.7 %