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Oxygen-15 undergoes beta decay. Which particle correctly completes the equation to show that the numbers of nucleons on the two sides of the equation are equal?

[tex]\[ _8^{15}O \rightarrow _7^{15}N + ? \][/tex]

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

B. [tex]\[ _1e \][/tex]

C. [tex]\[ ^3He \][/tex]

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

Sagot :

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

First, let's understand what beta decay is. Beta decay is a type of radioactive decay in which a neutron in the nucleus is converted into a proton while releasing an electron (beta particle). This electron is represented with the symbol [tex]\( _{-1}^{0} e \)[/tex].

Let's break down the given information:
The original nucleus is Oxygen-15: [tex]\( _{8}^{15}O \)[/tex]
After undergoing beta decay, it transforms into Nitrogen-15: [tex]\( _{7}^{15}N \)[/tex]

The challenge is to identify the correct particle that completes the equation so that the numbers of nucleons on both sides are equal:

[tex]\[ _{8}^{15}O \rightarrow _{7}^{15}N + ? \][/tex]

### Nucleon Number Analysis
1. The oxygen nucleus has 15 nucleons (total number of protons and neutrons combined).
2. The nitrogen nucleus still has 15 nucleons.

So, the sum of nucleon numbers on both sides should remain equal. The nucleon number of the missing particle is zero (since [tex]\( 15 = 15 + 0 \)[/tex]).

### Proton (Atomic) Number Analysis
1. The oxygen nucleus has 8 protons (atomic number 8).
2. The nitrogen nucleus has 7 protons (atomic number 7).

For charge conservation, the particle must account for the difference in proton numbers:
[tex]\[ 8 = 7 + ? \][/tex]

To balance the charges, the missing particle must have a charge of [tex]\( -1 \)[/tex] since [tex]\( 8 = 7 + (-1) \)[/tex].

### Conclusion
Combining these observations:
- The missing particle must have a nucleon number of 0.
- The missing particle must have an atomic number (charge) of -1.

The particle that matches these properties is the electron (beta particle), represented by [tex]\( _{-1}^{0}e \)[/tex].

Therefore, the correct answer is:

D. [tex]${ }_{-1}^0 e$[/tex]

This ensures that the numbers of nucleons and the charges on both sides of the equation are balanced.
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