Carbon-11 undergoes radioactive decay. Which particle correctly completes the equation to show that the numbers of nucleons on each side of the equation are equal?

[tex]\[ _6^{11}C \rightarrow _5^{11}B + ? \][/tex]

A. [tex]\[ _{-1}^0 e \][/tex]

B. [tex]\[ _4^2 He \][/tex]

C. [tex]\[ _{+1}^0 e \][/tex]

D. [tex]\[ _2^4 He \][/tex]



Answer :

Certainly! Let's solve this step-by-step.

First, it's important to understand the decay process at play here. We have Carbon-11 undergoing decay into Boron-11.

The general decay equation we are given is:
[tex]\[ _6^{11} C \rightarrow _5^{11} B + ? \][/tex]

We need to determine what particle completes this equation. To do this, we'll balance both the nucleon number (sum of protons and neutrons) and the proton (atomic) number on both sides of the equation.

Step 1: Balance the nucleon number

The initial carbon isotope [tex]\(_6^{11} C\)[/tex] has 11 nucleons (neutrons + protons). Boron-11 [tex]\(_5^{11} B\)[/tex] also has 11 nucleons, as indicated by the superscript.

Thus, the total nucleon number (11) is already balanced on both sides:
[tex]\[ _6^{11} C \rightarrow _5^{11} B + \text{(0 nucleons particle)} \][/tex]

Step 2: Balance the proton number

Carbon-11 has 6 protons as indicated by the subscript (atomic number = 6), while Boron-11 has 5 protons (atomic number = 5). This means we have:
[tex]\[ 6 \text{ protons (Carbon-11)} = 5 \text{ protons (Boron-11)} + \text{proton number of the missing particle} \][/tex]

To make both sides equal regarding the proton number, the missing particle must therefore have a single positive proton charge:
[tex]\[ 6 - 5 = 1 \][/tex]

So, we need a particle with a positive charge of +1 and 0 nucleons to satisfy this equation. The particle fitting this description is the positron, represented as [tex]\({ }_{+1}^0 e\)[/tex].

Thus, the correct particle to complete the equation is the positron:
[tex]\[ _6^{11} C \rightarrow _5^{11} B + _{+1}^0 e \][/tex]

Answer:

The particle that correctly completes the equation is:
C. [tex]\({ }_{+1}^0 e\)[/tex]