Q. 5.66 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the balanced nuclear equation for each of the following:$(5.2)$

a. Polonium-$210$ decays to give lead-$206$

b. Bismuth-$211$ emits an alpha particle.

c. A radioisotope emits a positron to form titanium-$48$.

d. An atom of germanium-$69$ emits a positron.

Step-by-Step Solution

Answer

  

(part a) ${}_{84}{}^{210}\mathrm{PO}\to {}_{82}{}^{206} \mathrm{Pb}+{}_2{}^4\mathrm{He}$

(part b) ${}_{83}{}^{211}\mathrm{Bi}\to {}_{81}{}^{207}\mathrm{Th}+{}_2{}^4\mathrm{He}$

(part c) ${}_{23}{}^{48}\mathrm{Zn}\to {}_{+1}{}^{6}e+{}_{22}{}^{45}\mathrm{Ti}$

(part d)${}_{32}{}^{69}\mathrm{Ge}\to {}_{31}{}^{69}\mathrm{Ga}+{}_{+1}{}^{0}e$

Step 1- Given Information (part a)

Both mass numbers and atomic numbers are balanced to balance the nuclear reaction. That is, the sum of the mass number and atomic number of the reactants is equal to the product.

Step 2 - Explanation (part a)

(part a) 

Use the following steps to determine the product. 

${}_{44}{}^{210}\mathrm{Po}\to {}_{x_2}{}^{206} \mathrm{Pb}+?$

Unknown product mass number: 

$210=A+206$

   $A=4$

Atomic number of an unknown product.

 $84=Z+82$

    $Z=2$

So the product is ${}_2{}^4 He$, and the complete reaction is: 

Step 3: Explanation (part b)

(part b)

Use the following steps to determine the product. 

${}_{81}{}^{21}\mathrm{Bi}\to ?+{}_2{}^4\mathrm{He}$

Unknown product mass number: 

$211=A+4$

   $A=207$

Atomic number of an unknown product.

$83=Z+2$

   $Z=81$

 So the product is $\begin{array}{r}207\\ 81\end{array}$ , and the complete reaction is: 

${}_{83}{}^{211}\mathrm{Bi}\to {}_{81}{}^{207}\mathrm{Th}+{}_2{}^4\mathrm{He}$

Step 4- Explanation (part c)

(part c)

Use the following steps to determine the product. 

$?\to {}_{+1}{}^{e}e+{}_{22}{}^{48}\mathrm{Ti}$

Unknown product mass number: 

$A=48+0$

    $=48$

Atomic number of an unknown product.

$Z=22+1$

$Z=23$

 So the product is ${}_{23}{}^{48}\mathrm{Zn}$ , and the complete reaction is: 

${}_{23}{}^{48}\mathrm{Zn}\to {}_{+1}{}^{6}e+{}_{22}{}^{45}\mathrm{Ti}$

Step 5- Explanation (part d)

(part d) 

Use the following steps to determine the product. 

${}_{32}{}^{\theta }\mathrm{Ge}\to ?+{}_{+1}{}^{e}e$

Unknown product mass number: 

   $69=A+0$

       $A=69$

Atomic number of an unknown product.

$32=Z+1$

    $Z=31$

 The product is ${}_{31}{}^{69}\mathrm{Ga}$ , and the complete reaction is: 

${}_{32}{}^{69}\mathrm{Ge}\to {}_{31}{}^{69}\mathrm{Ga}+{}_{+1}{}^{0}e$