Problem 69 - Chemistry The Central Science

Question

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Step-by-Step Solution

Answer

The stable nucleus formed after three alpha emissions and two beta emissions from Radon-222 is Lead-210 (Pb-210).

Step 1: Identify the initial element and its atomic number and mass number

The initial element is Radon-222 (Rn-222). For Rn-222, - Atomic number (Z) = 86 (number of protons) - Mass number (A) = 222 (number of protons + neutrons)

Step 2: Three alpha emissions

An alpha emission involves the loss of two protons and two neutrons from a nucleus. So, after each alpha emission, the atomic number will decrease by 2, and the mass number will decrease by 4. After the first alpha emission: - Atomic number = 86 - 2 = 84 - Mass number = 222 - 4 = 218 After the second alpha emission: - Atomic number = 84 - 2 = 82 - Mass number = 218 - 4 = 214 After the third alpha emission: - Atomic number = 82 - 2 = 80 - Mass number = 214 - 4 = 210 So at this point, we have a nucleus with an atomic number of 80 and a mass number of 210.

Step 3: Two beta emissions

A beta emission involves the conversion of a neutron into a proton while emitting an electron (beta particle). As a result, the atomic number will increase by 1 while the mass number remains the same. After the first beta emission: - Atomic number = 80 + 1 = 81 - Mass number = 210 After the second beta emission: - Atomic number = 81 + 1 = 82 - Mass number = 210 At this point, we have a nucleus with an atomic number of 82 and a mass number of 210.

Step 4: Identify the stable nucleus formed

The final stable nucleus has: - Atomic number (Z) = 82 - Mass number (A) = 210 This corresponds to the element Lead (Pb) with 82 protons and a mass number of 210. Therefore, the stable nucleus formed is Lead-210 (Pb-210).