Question

Complete and balance the following equations for nuclear reactions that are used in particle accelerators to make elements beyond uranium.

(a) ${}_{\mathbf{2}}{}^{\mathbf{4}}\mathbf{He}\mathbf{+}{}_{\mathbf{99}}{}^{\mathbf{253}}\mathbf{Es}\to \mathbf{?}\mathbf{+}\mathbf{2}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

(b) ${}_{\mathbf{98}}{}^{\mathbf{249}}\mathbf{Cf}\mathbf{+}\mathbf{?}\to {}_{\mathbf{103}}{}^{\mathbf{257}}\mathbf{Lr}\mathbf{+}\mathbf{2}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

(c) ${}_{\mathbf{92}}{}^{\mathbf{238}}\mathbf{U}\mathbf{+}{}_{\mathbf{6}}{}^{\mathbf{12}}\mathbf{C}\to {}_{\mathbf{98}}{}^{\mathbf{244}}\mathbf{Cf}\mathbf{+}\mathbf{?}$

Short answer

The complete balanced reactions for the nuclear reactions taking place in particle accelerators are given below.

(a) ${}_{\text{2}}{}^{\text{4}}\text{He +}{}_{\text{99}}{}^{\text{253}}\text{Es}\to {}_{\text{101}}{}^{\text{255}}\text{Md + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

(b) ${}_{\text{98}}{}^{\text{249}}\text{Cf +}{}_{\text{5}}{}^{\text{10}}\text{B}\to {}_{\text{103}}{}^{\text{257}}\text{Lr + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

(c)${}_{\text{92}}{}^{\text{238}}\text{U +}{}_{\text{6}}{}^{\text{12}}\text{C}\to {}_{\text{98}}{}^{\text{244}}\text{Cf + 6}{}_{\text{0}}{}^{\text{1}}\text{n}$

Step-by-step solution

  Nuclear reactions

In nuclear reactions, a heavier nucleus decays to form lighter particles while lighter nuclei combine to form heavier nucleus. The nuclear reactions are balanced by equating the sum of mass numbers and atomic numbers of reactants with products. Normally nuclear reactions are accompanied with various kinds of emission. The most common type of emissions are alpha particles, beta particles, and gamma radiations.

  Balancing reaction (a)

${}_{\text{2}}{}^{\text{4}}\text{He +}{}_{\text{99}}{}^{\text{253}}\text{Es}\to \text{? + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

To complete the reaction, we need to identify the missing product from the equation. It is evident from the equation that a heavier element is formed from the combination of two different lighter nuclei along with the liberation of 2 neutrons.

Sum of mass number of the reactants $=253+4=257$

Sum of atomic number of the reactants$=99+2=101$

Therefore, mass number of the missing product $=257-2=255$

And, atomic number of the missing product$=101$

The element is ${}_{101}{}^{255}\text{Md}$.

The complete balanced reaction is given below.

${}_{\text{2}}{}^{\text{4}}\text{He +}{}_{\text{99}}{}^{\text{253}}\text{Es}\to {}_{\text{101}}{}^{\text{255}}\text{Md + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

  Balancing reaction (b)

${}_{\text{98}}{}^{\text{249}}\text{Cf + ?}\to {}_{\text{103}}{}^{\text{257}}\text{Lr + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

In this reaction we have to identify the missing reactant.

The sum of mass number of the reactant $=257+2=259$

The sum of atomic number of the reactant$=103$

The mass number of the missing reactant$=259-249=10$

The atomic number of the missing reactant$=103-98=5$

The element is ${}_{\text{5}}{}^{\text{10}}\text{B}$.

The complete balanced reaction is given below.

${}_{\text{98}}{}^{\text{249}}\text{Cf +}{}_{\text{5}}{}^{\text{10}}\text{B}\to {}_{\text{103}}{}^{\text{257}}\text{Lr + 2}{}_{\text{0}}{}^{\text{1}}\text{n}$

  Balancing reaction (c)

${}_{\text{92}}{}^{\text{238}}\text{U +}{}_{\text{6}}{}^{\text{12}}\text{C}\to {}_{\text{98}}{}^{\text{244}}\text{Cf + ?}$

In this reaction, two nuclei combine to form a heavier nucleus. The reaction is accompanied by the emission of neutrons. We have to calculate the difference in the mass number of the reactants and products to find out the number of neutrons emitted.

Sum of mass number of reactants$=238+12=250$

Sum of atomic number of reactants$=92+6=98$

Number of neutrons$=250-244=6$

The complete balanced equation is given below.

${}_{\text{92}}{}^{\text{238}}\text{U +}{}_{\text{6}}{}^{\text{12}}\text{C}\to {}_{\text{98}}{}^{\text{244}}\text{Cf + 6}{}_{\text{0}}{}^{\text{1}}\text{n}$