The emission of two successive beta particles from the nucleus $^{32} _{15} P$ will produce

Analysis of the Question

This question tests your understanding of radioactive decay, specifically beta ($\beta$) emission. You are asked to determine the resulting nucleus after a phosphorus-32 atom undergoes two successive beta decays.

Concept Explanation

A beta particle is essentially a high-speed electron ($^{0}_{-1}e$). When a nucleus emits a beta particle:

1. Mass Number ($A$): Remains unchanged because the mass of an electron is negligible.
2. Atomic Number ($Z$): Increases by 1. This happens because a neutron in the nucleus converts into a proton and an electron (the beta particle), which is then ejected.

The Rule: If a nucleus $^{A}_{Z}X$ emits one beta particle: $^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}e$

Step-by-Step Solution

Step 1: Identify the initial nucleus. The starting material is $^{32}_{15}P$.

• Mass number ($A$) = 32
• Atomic number ($Z$) = 15

Step 2: First beta emission. $^{32}_{15}P \rightarrow ^{32}_{15+1}X + ^{0}_{-1}e$ $^{32}_{15}P \rightarrow ^{32}_{16}S + ^{0}_{-1}e$ (The element with atomic number 16 is Sulfur, S).

Step 3: Second beta emission. Now, the new nucleus ($^{32}_{16}S$) undergoes a second beta decay: $^{32}_{16}S \rightarrow ^{32}_{16+1}Y + ^{0}_{-1}e$ $^{32}_{16}S \rightarrow ^{32}_{17}Cl + ^{0}_{-1}e$ (The element with atomic number 17 is Chlorine, Cl).

Conclusion

After two successive beta emissions, the atomic number increases by 2 (from 15 to 17), while the mass number remains 32. The resulting nucleus is $^{32}_{17}Cl$.

Comparing this to the options provided: A. $^{30} _{14}$Si B. $^{32} _{13}$Al C. $^{32} _{17}$Cl D. $^{32} _{16}$S

The correct option is C.

Exam Tip

Remember the "Beta-Plus-One" rule: Every time you see "beta decay," just add 1 to the atomic number and keep the mass number the same. If it says "two successive," just add 2 to the atomic number. Don't let the "32" confuse you—beta decay never changes the mass number!