Q. 5.65 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. When two oxygen-$16$ atoms collide, one of the products is an alpha particle.

b. When californium- $249$is bombarded by oxygen- $18$, a new element, seaborgium-$263$, and four neutrons are produced.

c. Radon- $222$ undergoes alpha decay.

d. An atom of strontium- $80$ emits a positron.

Step-by-Step Solution

Answer

The core equation is calculated 

a. ${}_8{}^{16}\mathrm{O}+{}_8{}^{16}\mathrm{O}\to {}_{14}{}^{28}\mathrm{Si}+{}_2{}^4\mathrm{He}$

b.${}_8{}^{18}\mathrm{O}+{}_{98}{}^{249}\mathrm{Cf}\to {}_{106}{}^{263}\mathrm{Cf}+4{}_0{}^{1}n$

c. ${}_{86}{}^{222}\mathrm{Ra}\to {}_{84}{}^{218}\mathrm{PO}+{}_2{}^4\mathrm{He}$

d.${}_{38}{}^{80}\mathrm{Sr}\to {}_{37}{}^{80}\mathrm{Rb}+{}_{+1}{}^{0}e$

Step 1- Given Information

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- Given explanation (part a)

a.

Use the following steps to determine the product.

${}_8{}^{16}\mathrm{O}+{}_8{}^{16}\mathrm{O}\to ?+{}_2{}^4\mathrm{He}$

Unknown product mass number:

$16+16=A-4$

     $A=28$

Atomic number of unknown product: 

   $8+8=Z-2$

    $Z=14$

So the product is ${}_{14}{}^{28} Si$, and the complete reaction is:

 ${}_8{}^{16}\mathrm{O}+{}_8{}^{16}\mathrm{O}\to {}_{14}{}^{28}\mathrm{Si}+{}_2{}^4\mathrm{He}$

Step 3- Explanation (part b)

b.

Use the following steps to determine the product.

${}_8{}^{18}\mathrm{O}+{}^{200}\mathrm{Cf}\to ?+4{}_0{}^{1}n$

Unknown product mass number:

$168+249=A-4\times 1$

     $A=263$

Atomic number of unknown product: 

   $8+98=Z-0$

    $Z=106$

So the product is ${}_{106}{}^{263}\mathrm{Cf}$   , and the complete reaction is:

 ${}_8{}^{18}\mathrm{O}+{}_{98}{}^{249}\mathrm{Cf}\to {}_{106}{}^{263}\mathrm{Cf}+4{}_0{}^{1}n$

Step 4- Explanation (part c)

c.

Use the following steps to determine the product.

${}_{36}{}^{222}\mathrm{Ra}\to ?+{}_2{}^4\mathrm{He}$

Unknown product mass number:

$222=A-4$

   $A=218$ 

Atomic number of unknown product: 

   $86=Z-2$

    $Z=84$

So the product is ${}_{84}{}^{218}\mathrm{PO}$  , and the complete reaction is:

${}_{86}{}^{222}\mathrm{Ra}\to {}_{84}{}^{218}\mathrm{PO}+{}_2{}^4\mathrm{He}$

Step 5- Explanation (part d)

d.

Use the following steps to determine the product.

${}_{38}{}^{80}\mathrm{Sr}\to ?+{}_{+1}{}^{0}e$

Unknown product mass number:

$80=A+0$

    $A=80$

Atomic number of unknown product: 

   $38=Z+1$

    $Z=37$

The product is ${}_{37}{}^{80}\mathrm{Rb}$ , and the complete reaction is  

${}_{38}{}^{80}\mathrm{Sr}\to {}_{37}{}^{80}\mathrm{Rb}+{}_{+1}{}^{0}e$