Radon (86Rn222) is an unstable isotope that undergoes α decay with a half-life of3.8 days.(i) Identify the daughter nucleus in this decay process. Show how you obtained youranswer. [2 marks](ii) Calculate the activity of a sample containing 2.0 x 108 atoms of 86Rn222.[3 marks](iii) Estimate how many α particles are emitted in the first hour after the sample isformed.

(i) In alpha decay, a nucleus emits an alpha particle, which is a helium nucleus (2 protons and 2 neutrons). Therefore, the atomic number decreases by 2 and the mass number decreases by 4. For Radon (86Rn222), the daughter nucleus would be Polonium (84Po218). Here's the equation:

86Rn222 → 84Po218 + 2He4

(ii) The activity of a radioactive sample is given by the equation A = λN, where λ is the decay constant and N is the number of atoms. The decay constant can be found using the equation λ = ln(2) / T, where T is the half-life.

First, convert the half-life to seconds: 3.8 days = 3.8 * 24 * 60 * 60 = 328320 seconds.

Then, calculate the decay constant: λ = ln(2) / 328320 = 2.11 x 10^-6 s^-1.

Finally, calculate the activity: A = λN = 2.11 x 10^-6 s^-1 * 2.0 x 10^8 atoms = 422 Bq.

(iii) The number of alpha particles emitted in a given time is given by the equation N = N0 * (1 - e^-λt), where N0 is the initial number of atoms, λ is the decay constant, and t is the time.

First, convert the time to seconds: 1 hour = 60 * 60 = 3600 seconds.

Then, calculate the number of alpha particles: N = 2.0 x 10^8 atoms * (1 - e^-2.11 x 10^-6 s^-1 * 3600 s) ≈ 2.7 x 10^6 alpha particles.