Q24.53 P - Chemistry: Molecular Nature Of Matter And Change

Question

Determine the missing species in these transmutations, and write a full nuclear equation from the shorthand notation:

_{${\mathbf{\text{ (a) }}}^{\mathbf{\text{10}}}\mathbf{\text{ B(α,n)}}$}

$\begin{array}{l}{\mathbf{\text{(b)}}}^{\mathbf{\text{28}}}{\mathbf{\text{Si(d,)}}}^{\mathbf{\text{29}}}{\mathbf{\text{P(whered is a duteron,}}}^{\mathbf{\text{2}}}{\mathbf{\text{H}}}^{\mathbf{\text{+}}}\mathbf{\text{)}}\\ {\mathbf{\text{(c)(α,2n)}}}^{\mathbf{\text{244}}}\mathbf{\text{Cf}}\end{array}$

Step-by-Step Solution

Answer

$\begin{array}{l}{\text{a)}}_{\text{5}}^{\text{10}}{\text{ B+}}_{\text{2}}^{\text{4}}\text{He}{\to }_{\text{0}}^{\text{1}}{\text{n+}}_{\text{7}}^{\text{13}}\text{ N}\\ {\text{b)}}_{\text{14}}^{\text{8}}{\text{Si+}}_{\text{1}}^{\text{2}}\text{H}{\to }_{\text{0}}^{\text{1}}{\text{n+}}_{\text{15}}^{\text{29}}\text{P}\\ {\text{c)}}_{\text{6}}^{\text{242}}{\text{Cm+}}_{\text{2}}^{\text{4}}\text{He}\to {\text{2}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{08}}^{\text{244}}\text{Cf}\end{array}$

Step 1: Subpart (a)

- A: mass number

- Z: atomic number

- Let X be the unknown element or particle

(a)  _{${}^{\text{10}}\text{B(α,n)}$}

- The atomic number of B is 5

- Alpha particle:  _{${}_2^4\text{He}$}

- Neutron: ${}_{\text{0}}^{\text{1}}\text{n}$ 

 _{${}_{\text{5}}^{\text{10}}{\text{ B+}}_{\text{2}}^{\text{4}}\text{He}{\to }_{\text{0}}^{\text{1}}{\text{n+}}_{\text{Z}}^{\text{A}}\text{X}$}

First, let us calculate the values of A and Z

 _{$\text{10+4 =1+AA =135+2 =0+ZZ =7}$}

Hence The element with atomic number  7 is nitrogen, so the product is  _{${}_7^{13} \text{N}$}, and the balanced equation is

 _{${}_{\text{5}}^{\text{10}}{\text{ B+}}_{\text{2}}^{\text{4}}\text{He}{\to }_{\text{0}}^{\text{1}}{\text{n+}}_{\text{7}}^{\text{13}}\text{ N}$}

Step 2: Subpart (b)

 _{${}^{28}\text{Si}(\text{d}{,}^{29}\text{P}$}

- The atomic number of Si is  $14$

- The atomic number of P is  _$15$

- deuteron (d): ${}_1^2\text{H}$ 

 _{${}_{14}^{28}\text{Si}{+}_1^2\text{H}{\to }_{\text{Z}}^{\text{A}}\text{X}{+}_{15}^{29}\text{P}$}

First, let us calculate the values of A and Z

 _{$\text{28+2 =A+29 A =114+1 =Z+15Z =0}$}

Hence The particle with mass number 1 , and charge 0 is neutron

 _{${}_{14}^{28}\text{Si}{+}_1^2\text{H}{\to }_0^1\text{n}{+}_{15}^{29}\text{P}$}

Step 3: Subpart (c)

 _{${\text{(α,2n)}}^{\text{244}}\text{Cf}$}

- The atomic number of Cf is $98$  

- Alpha particle:  _{data-custom-editor="chemistry" ${}_2^4\text{He}$}

- Neutron:  _{${}_{\text{0}}^{\text{1}}\text{n}$}

 _{${}_{\text{Z}}^{\text{A}}{\text{X+}}_{\text{2}}^{\text{4}}\text{He}\to {\text{2}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{98}}^{\text{244}}\text{Cf}$}

First, let us calculate the values of A and Z

 _{data-custom-editor="chemistry" $\text{A+4 =2×1+244 A =242Z+2 =2×0+98Z =96}$}

Hence, the element with atomic number 96 is curium, so the reactant is  _{${}_{96}^{242}\text{Cm}$}, and the balanced equation is ${}_{\text{96}}^{\text{242}}{\text{Cm+}}_{\text{2}}^{\text{4}}\text{He}\to {\text{2}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{98}}^{\text{244}}\text{Cf}$