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

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

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

- 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

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

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

- 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

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

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

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

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

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

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