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1 0 - N u m e r i c a l   7

Problem

Complete the following nuclear reactions:

7N14+2He4β†’1H1+?5B11+1H1β†’6C11+?3Li6+?β†’4Be7+0n1\begin{aligned} & {}_{7}\mathrm{N}^{14} + {}_{2}\mathrm{He}^{4} \rightarrow {}_{1}\mathrm{H}^{1} + ? \\ & {}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{1} \rightarrow {}_{6}\mathrm{C}^{11} + ? \\ & {}_{3}\mathrm{Li}^{6} + ? \rightarrow {}_{4}\mathrm{Be}^{7} + {}_{0}\mathrm{n}^{1} \end{aligned}

Data

  • Reaction 1:
7N14+2He4β†’1H1+? {}_{7}\mathrm{N}^{14} + {}_{2}\mathrm{He}^{4} \rightarrow {}_{1}\mathrm{H}^{1} + ?

Known products: Proton (1H1{}_{1}\mathrm{H}^{1})
Unknown product to be determined.

  • Reaction 2:
5B11+1H1β†’6C11+? {}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{1} \rightarrow {}_{6}\mathrm{C}^{11} + ?

Known products: Carbon-11 (6C11{}_{6}\mathrm{C}^{11})
Unknown product to be determined.

  • Reaction 3:
3Li6+?β†’4Be7+0n1 {}_{3}\mathrm{Li}^{6} + ? \rightarrow {}_{4}\mathrm{Be}^{7} + {}_{0}\mathrm{n}^{1}

Known products: Beryllium-7 (4Be7{}_{4}\mathrm{Be}^{7}), Neutron (0n1{}_{0}\mathrm{n}^{1})
Unknown reactant to be determined.

Prerequisite Concepts

  1. Balancing Nuclear Reactions:
    The sum of mass numbers (AA) and atomic numbers (ZZ) must be equal on both sides of the reaction.

  2. Notation:

    • AA: Mass number (total protons and neutrons)
    • ZZ: Atomic number (number of protons)
  3. Reaction Symbols:

    • Alpha (Ξ±\alpha): 2He4{}_{2}\mathrm{He}^{4}
    • Proton (pp): 1H1{}_{1}\mathrm{H}^{1}
    • Neutron (nn): 0n1{}_{0}\mathrm{n}^{1}
    • Deuteron (dd): 1H2{}_{1}\mathrm{H}^{2}

Solution

Reaction 1:

7N14+2He4β†’1H1+?{}_{7}\mathrm{N}^{14} + {}_{2}\mathrm{He}^{4} \rightarrow {}_{1}\mathrm{H}^{1} + ?
  1. Initial reactants:
7N14(Nitrogen-14),2He4(AlphaΒ particle) {}_{7}\mathrm{N}^{14} \quad \text{(Nitrogen-14)}, \quad {}_{2}\mathrm{He}^{4} \quad \text{(Alpha particle)}
  1. Known product:
1H1(Proton) {}_{1}\mathrm{H}^{1} \quad \text{(Proton)}
  1. Determine the unknown product:
    Balance atomic and mass numbers:

    • Mass: 14+4=18β†’1+1714 + 4 = 18 \rightarrow 1 + 17
    • Atomic: 7+2=9β†’1+87 + 2 = 9 \rightarrow 1 + 8

    Unknown product: 8O17(Oxygen-17){}_{8}\mathrm{O}^{17} \quad \text{(Oxygen-17)}

    Complete reaction:

7N14+2He4β†’1H1+8O17 {}_{7}\mathrm{N}^{14} + {}_{2}\mathrm{He}^{4} \rightarrow {}_{1}\mathrm{H}^{1} + {}_{8}\mathrm{O}^{17}

Reaction 2:

5B11+1H1β†’6C11+?{}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{1} \rightarrow {}_{6}\mathrm{C}^{11} + ?
  1. Initial reactants:
5B11(Boron-11),1H1(Proton) {}_{5}\mathrm{B}^{11} \quad \text{(Boron-11)}, \quad {}_{1}\mathrm{H}^{1} \quad \text{(Proton)}
  1. Known product:
6C11(Carbon-11) {}_{6}\mathrm{C}^{11} \quad \text{(Carbon-11)}
  1. Determine the unknown product:
    Balance atomic and mass numbers:

    • Mass: 11+1=12β†’11+111 + 1 = 12 \rightarrow 11 + 1
    • Atomic: 5+1=6β†’6+05 + 1 = 6 \rightarrow 6 + 0

    Unknown product: 0n1(Neutron){}_{0}\mathrm{n}^{1} \quad \text{(Neutron)}

    Complete reaction:

5B11+1H1β†’6C11+0n1 {}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{1} \rightarrow {}_{6}\mathrm{C}^{11} + {}_{0}\mathrm{n}^{1}

Reaction 3:

3Li6+?β†’4Be7+0n1{}_{3}\mathrm{Li}^{6} + ? \rightarrow {}_{4}\mathrm{Be}^{7} + {}_{0}\mathrm{n}^{1}
  1. Known products:
4Be7(Beryllium-7),0n1(Neutron) {}_{4}\mathrm{Be}^{7} \quad \text{(Beryllium-7)}, \quad {}_{0}\mathrm{n}^{1} \quad \text{(Neutron)}
  1. Initial reactant:
3Li6(Lithium-6) {}_{3}\mathrm{Li}^{6} \quad \text{(Lithium-6)}
  1. Determine the unknown reactant:
    Balance atomic and mass numbers:

    • Mass: 6+?=7+1β†’?=26 + ? = 7 + 1 \rightarrow ? = 2
    • Atomic: 3+?=4+0β†’?=13 + ? = 4 + 0 \rightarrow ? = 1

    Unknown reactant: 1H2(Deuteron){}_{1}\mathrm{H}^{2} \quad \text{(Deuteron)}

    Complete reaction:

3Li6+1H2β†’4Be7+0n1 {}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow {}_{4}\mathrm{Be}^{7} + {}_{0}\mathrm{n}^{1}

Answer

  1. 7N14+2He4β†’1H1+8O17{}_{7}\mathrm{N}^{14} + {}_{2}\mathrm{He}^{4} \rightarrow {}_{1}\mathrm{H}^{1} + {}_{8}\mathrm{O}^{17}
  2. 5B11+1H1β†’6C11+0n1{}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{1} \rightarrow {}_{6}\mathrm{C}^{11} + {}_{0}\mathrm{n}^{1}
  3. 3Li6+1H2β†’4Be7+0n1{}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow {}_{4}\mathrm{Be}^{7} + {}_{0}\mathrm{n}^{1}