Q. 5.62 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the balanced nuclear equation for each of the following: $(5.1,5.2)$

a. potassium- $40$ ( $\beta$ decay)

b. sulfur-$35$( $\beta$ decay)

c. platinum-$190$  $(\alpha$ decay)

d.   Ra- $210$ ( $\alpha$ decay)

e. In- $113m$ ( $\gamma$ emission)

Step-by-Step Solution

Answer

The balanced nuclear equations are derived below, 

(part a)${{}_{19} \mathrm{K}}^{40}\to {{}_{20}\mathrm{Ca}}^{40}+{{}_{-1}\mathrm{e}}^0$

(part b)${{}_{16} \mathrm{S}}^{35}\to {{}_{17}\mathrm{Cl}}^{35}+{{}_{-1}\mathrm{e}}^0$

(part c)  ${{}_{78}\mathrm{Pt}}^{190}\to {{}_{76}\mathrm{Os}}^{186}+{{}_2\mathrm{He}}^4$

(part d)   ${{}_{88}\mathrm{Ra}}^{210}\to {{}_{86}\mathrm{Rn}}^{206}+{{}_2\mathrm{He}}^4$

(part e)  ${{}_{49}In}^{113 \mathrm{m}}\to {{}_{49}In}^{113}+{{}_0\gamma }^0$

Step 1 - Given information (part a)

(part a) 

 ${{}_{19} \mathrm{K}}^{40}\to ?+{{}_{-1}\mathrm{e}}^0$

The mass number of potassium is equal to the sum of the mass numbers of beta particles and  new nuclei.

$40=?+0$

$40= Mass number of New Nucleus$

The atomic number of potassium $-40$must be equal to the sum of the atomic numbers of the beta particles and the new nuclei.

$19=?+(-1)$

$19+1=?$

$20= Atomic number of New Nucleus$

The image of the isotope is ${{}_{20}\mathrm{Ca}}^{40}$

${{}_{19} \mathrm{K}}^{40}\to {{}_{20}\mathrm{Ca}}^{40}+{{}_{-1}\mathrm{e}}^0$

Step 2- Given information (Part b)

(part b)

${{}_{16} \mathrm{S}}^{35}\to ?+{{}_{-1}\mathrm{e}}^0$

The mass number of is equal to the sum of the mass numbers of the beta particles and the new nuclei.

$35=?+0$

$35= Mass number of New Nucleus$

The atomic number of $35$ must be equal to the sum of the atomic numbers of the beta particles and the new nuclei.

$16=?+(-1)$

$16+1=?$

$17= Atomic number of New Nucleus$

The image of the isotope is ${{}_{17}\mathrm{Cl}}^{35}$

${{}_{16} \mathrm{S}}^{35}\to {{}_{17}\mathrm{Cl}}^{35}+{{}_{-1}\mathrm{e}}^0$

Step 3- Explanation -1 (part a)

(part a) 

${{}_{78}\mathrm{Pt}}^{190}\to ?+{{}_2\mathrm{He}}^4$

The mass number of platinum is equal to the sum of the mass numbers of the alpha particles and the new nuclei.

width="78" style="max-width: none; vertical-align: -4px;" $190=?+4$

$190-4=?$

$186= Mass number of New Nucleus$

The atomic number of Platinum $190$ must be equal to the sum of the atomic numbers of the alpha particles and the new nuclei.

$78=?+4$

$78-2=?$

width="273" style="max-width: none; vertical-align: -5px;" $76= Atomic number of New Nucleus$

The image of the isotope is ${{}_{76}\mathrm{Os}}^{186}$

${{}_{78}\mathrm{Pt}}^{190}\to {{}_{76}\mathrm{Os}}^{186}+{{}_2\mathrm{He}}^4$

Step 4- Explanation -1 (part b)

(part b)

${{}_{88}\mathrm{Ra}}^{210}\to ?+{{}_2\mathrm{He}}^4$

The mass number of radium is equal to the sum of the mass numbers of alpha particles and  new nuclei.

$210=?+4$

$210-4=?$

$206=Mass number of New Nucleus$

The atomic number of radium $210$ is equal to the sum of the atomic numbers of the alpha particles and the new nuclei.

$88=?+2$

$88-2=?$

$86= Atomic number of New Nucleus$

The image of the isotope is ${{}_{86}\mathrm{Rn}}^{206}$

${{}_{88}\mathrm{Ra}}^{210}\to {{}_{86}\mathrm{Rn}}^{206}+{{}_2\mathrm{He}}^4$

Step 5- Explanation 1 (part c)

(part c) 

${{}_{49}In}^{113 \mathrm{m}}\to ?+{{}_0\gamma }^0$

Like the reaction above, it involves the release of $\gamma -emission.$  . There is  no net charge on the atomic number and atomic weight of the nucleus.

$m$ indicates the atomically stable state of indium 

    The image of the isotope is ${{}_{49}In}^{113}$

${{}_{49}In}^{113 \mathrm{m}}\to {{}_{49}In}^{113}+{{}_0\gamma }^0$