Question

1. radon - 221 decays in a series of steps emitting four alpha and three beta particles to become a stable isotope. what is the identity of this stable isotope? support your answer by showing a series of steps or one overall summary equation. (4pts)

Explanation:

Step1: Recall alpha - decay equation

An alpha - particle is a helium nucleus with mass number $A = 4$ and atomic number $Z=2$. The general form of alpha - decay is $_{Z}^{A}X
ightarrow_{Z - 2}^{A - 4}Y+_{2}^{4}He$. A beta - particle is an electron with mass number $A = 0$ and atomic number $Z=-1$. The general form of beta - decay is $_{Z}^{A}X
ightarrow_{Z + 1}^{A}Y+_{- 1}^{0}e$.

Step2: Calculate the change in mass number due to alpha - decay

Since there are 4 alpha - decays, the total decrease in mass number $\Delta A_{alpha}=4\times4 = 16$.

Step3: Calculate the change in atomic number due to alpha - decay

The total decrease in atomic number due to 4 alpha - decays is $\Delta Z_{alpha}=4\times2 = 8$.

Step4: Calculate the change in atomic number due to beta - decay

Since there are 3 beta - decays, the total increase in atomic number $\Delta Z_{beta}=3\times1=3$.

Step5: Determine the initial mass number and atomic number of Radon - 221

The atomic number of Radon (Rn) is $Z = 86$ and the mass number $A = 221$.

Step6: Calculate the final mass number

The final mass number $A_{final}=221-16=205$.

Step7: Calculate the final atomic number

The final atomic number $Z_{final}=86 - 8+3=81$.
The element with atomic number $Z = 81$ is Thallium (Tl).

Answer:

The stable isotope is Thallium - 205 ($_{81}^{205}Tl$)