Question

p⁺= e⁻= n⁰=

1. determine the number of protons, neutrons, and electrons of the hydrogen isotopes.

¹₁h ²₁h = deuterium ³₁h = tritium
p⁺=
e⁻=
n⁰=
|p⁺=
e⁻=
n⁰= |p⁺=
e⁻=
n⁰= |p⁺=
e⁻=
n⁰= |

Explanation:

Step1: Recall isotope notation

For an isotope \(_Z^A X\), \(Z\) is atomic number (protons, \(p^+\)), \(A\) is mass number (protons + neutrons, \(n^0\)). Electrons (\(e^-\)) equal protons in neutral atoms.

Step2: Analyze \(_1^1 H\)

• \(p^+\): \(Z = 1\) → \(p^+ = 1\)
• \(e^-\): Neutral → \(e^- = 1\)
• \(n^0\): \(A - Z = 1 - 1 = 0\)

Step3: Analyze \(_1^2 H\) (Deuterium)

• \(p^+\): \(Z = 1\) → \(p^+ = 1\)
• \(e^-\): Neutral → \(e^- = 1\)
• \(n^0\): \(A - Z = 2 - 1 = 1\)

Step4: Analyze \(_1^3 H\) (Tritium)

• \(p^+\): \(Z = 1\) → \(p^+ = 1\)
• \(e^-\): Neutral → \(e^- = 1\)
• \(n^0\): \(A - Z = 3 - 1 = 2\)

Answer:

For \(\boldsymbol{_1^1 H}\):
\(p^+ = 1\), \(e^- = 1\), \(n^0 = 0\)

For \(\boldsymbol{_1^2 H}\) (Deuterium):
\(p^+ = 1\), \(e^- = 1\), \(n^0 = 1\)

For \(\boldsymbol{_1^3 H}\) (Tritium):
\(p^+ = 1\), \(e^- = 1\), \(n^0 = 2\)