Chapter 25, Problem 95A

Interpretation Introduction

Interpretation: The balanced equations for two nuclear reactions are to be written.

Concept Introduction: When an atomic nucleus or more atomic nuclei and a subatomic particle collide, one or more nuclides also known as nuclear reactions are created.

Answer to Problem 95A

The two complete nuclear reaction is as follows:

  ${}_{\text{109}}^{270}\text{Mt}{\to }_{\text{107}}^{266}\text{Bh}{+}_{\text{2}}^{\text{4}}\text{He}$

  ${}_{\text{107}}^{266}\text{Bh}{\to }_{105}^{262}\text{Db}{+}_{\text{2}}^{\text{4}}\text{He}$

Explanation of Solution

According to nuclear reactions, the sum of the atomic number and mass number on the reactant side is equal to the sum of the atomic number and mass on the product side.

Element 107 $\left(\text{Bh}\right)$ is formed when nuclei of element 109 $\left(\text{Mt}\right)$ each emit an alpha particle.

Let $m$ and $n$ be the mass number of element 107 and element 109 respectively.

  ${}_{\text{109}}^{\text{m}}\text{Mt}{\to }_{\text{107}}^{\text{n}}\text{Bh}{+}_{\text{2}}^{\text{4}}\text{He}$   ...... (1)

Nuclei of element 107 each emit an alpha particle forming an atom with a mass number of 262.

Let $\text{Y}$ be element formed from element 107 and $x$ be its atomic number.

  ${}_{\text{107}}^{\text{n}}\text{Bh}{\to }_x^{262}\text{Y}{+}_{\text{2}}^{\text{4}}\text{He}$   ...... (2)

Equate the values of the atomic number of the reactants and products.

  $\begin{array}{l}107=x+2\\ x=107-2\\ x=105\end{array}$

Equate the values of the mass number of the reactants and products.

  $\begin{array}{l}n=262+4\\ n=266\end{array}$

The element with atomic number 105 is Dubnium.

The equation (2) is rewritten after the substitution of values.

  ${}_{\text{107}}^{266}\text{Bh}{\to }_{105}^{262}\text{Db}{+}_{\text{2}}^{\text{4}}\text{He}$

The equation (1) is rewritten after the substitution of values.

  ${}_{\text{109}}^{\text{m}}\text{Mt}{\to }_{\text{107}}^{266}\text{Bh}{+}_{\text{2}}^{\text{4}}\text{He}$   ...... (3)

Equate the values of mass number of the reactants and products.

  $\begin{array}{l}m=266+4\\ m=270\end{array}$

The equation (3) is rewritten after the substitution of values.

  ${}_{\text{109}}^{270}\text{Mt}{\to }_{\text{107}}^{266}\text{Bh}{+}_{\text{2}}^{\text{4}}\text{He}$