Complete the following nuclear reactions. \begin{equation}a. _{5}^{12} \mathrm{B} \rightarrow_{6}^{12} \mathrm{C}+?\end{equation} \begin{equation}b. _{89}^{225} \mathrm{Ac} \longrightarrow_{87}^{221} \mathrm{Fr}+?\end{equation} \begin{equation}\mathbf{c.}_{28}^{63} \mathrm{Ni} \longrightarrow ?+_{-1}^{0} e\end{equation} \begin{equation}\mathbf{d} \cdot_{83}^{212} \mathrm{Bi} \rightarrow ?+_{2}^{4} \mathrm{He}\end{equation}

Short Answer

Expert verified
The missing components for the reactions are: a) a Beta particle b) an alpha particle c) Copper nucleus d) Thallium nucleus

Step by step solution

01

Completing Reaction a

Here we observe that the atomic number changes from 5 to 6, hence we must have lost a -1 charge, denoting a beta particle. The reaction can be written as \( _{5}^{12} B → _{6}^{12} C + _{-1}^{0}\beta \)
02

Completing Reaction b

The atomic number decreases by 2 and the mass number decreases by 4. Hence, the missing particle in this case is an alpha particle. The reaction can be written as \( _{89}^{225} Ac → _{87}^{221} Fr + _{2}^{4}\alpha \)
03

Completing Reaction c

In this reaction, there is an emission of an electron particle. Therefore, the atomic number must increase by one. The resultant nucleus is thus named Copper. Hence the reaction can be written as \( _{28}^{63} Ni → _{29}^{63} Cu + _{-1}^{0} e \)
04

Completing Reaction d

Just like in step 2, the emission of the alpha particle means a reduction in atomic number by 2 and mass number by 4. Hence, the missing nucleus is Thallium. The reaction can be written as \( _{83}^{212} Bi → _{81}^{208} Tl + _{2}^{4} He \)

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