Q. 5.20 - An Introduction to General, Organic, and Biological Chemistry

Question

Complete each of the following nuclear equations and describe the type of radiation:

(a)${}_6{}^{11}\mathrm{C}\to {}_5{}^{11}\mathrm{B}+?$

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

(c)$?\to {}_{39}{}^{90}\mathrm{Y}+{}_{-1}{}^0\mathrm{e}$

(d)${}_{83}{}^{210}\mathrm{Bi}\to ?+{}_2{}^4\mathrm{He}$

(e)$?\to {}_{39}{}^{89}\mathrm{Y}+{}_1{}^0\mathrm{e}$

Step-by-Step Solution

Answer

(a) The complete balanced nuclear equation is${}_6{}^{11}\mathrm{C}\to {}_5{}^{11}\mathrm{B}+{}_1{}^0\mathrm{e}$ . It is positron emission.

(b) The complete balanced nuclear equation is${}_{16}{}^{35}\mathrm{S}\to {}_{17}{}^{35}Cl+{}_{-1}{}^0\mathrm{e}$. It is beta decay.

(c) The complete balanced nuclear equation is${}_{38}{}^{90}\mathrm{Sr}\to {}_{39}{}^{90}\mathrm{Y}+{}_{-1}{}^0\mathrm{e}$. It is beta decay.

(d) The complete balanced nuclear equation is${}_{83}{}^{210}\mathrm{Bi}\to {}_{81}{}^{210}\mathrm{Tl}+{}_2{}^4\mathrm{He}$. It is alpha decay.

(e) The complete balanced nuclear equation is${}_{40}{}^{89}\mathrm{Zr}\to {}_{39}{}^{89}\mathrm{Y}+{}_1{}^0\mathrm{e}$. It is positron emission.  

Part (a) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (a) Step 2: Explanation

To complete the equation,

${}_6{}^{11}\mathrm{C}\to {}_5{}^{11}\mathrm{B}+?$,

equate the mass number,

We get$11=11+?$,

So, the mass number is$0$ .

Now, equating the atomic number,

We get$6=5+?$,

So, the atomic number is$1$.

So, the element is e.

Hence, the equation is${}_6{}^{11}\mathrm{C}\to {}_5{}^{11}\mathrm{B}+{}_1{}^0\mathrm{e}$ and as the atomic number is decreasing by one, it is positron emission.

Part (b) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (b) Step 2: Explanation

To complete the equation,

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

equate the mass number,

We get$35=?+0$,

So, the mass number$35$ .

Now, equating the atomic number ,

We get$16=?-1$,

So, the atomic number is$17$.

So, the element is chlorine .

Hence, the equation is ${}_{16}{}^{35}\mathrm{S}\to {}_{17}{}^{35}Cl+{}_{-1}{}^0\mathrm{e}$ and as the atomic number is increasing by one, it is beta decay.

Part (c) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (c) Step 2: Explanation

To complete the equation,

$?\to {}_{39}{}^{90}\mathrm{Y}+{}_{-1}{}^0\mathrm{e}$,

equate the mass number,

We get$?=90+0$,

So, the mass number is$90$.

Now, equating the atomic number ,

We get$?=39-1$,

So, the atomic number is$38$

So, the element is strontium.

Hence, the equation is${}_{38}{}^{90}\mathrm{Sr}\to {}_{39}{}^{90}\mathrm{Y}+{}_{-1}{}^0\mathrm{e}$ and as the atomic number is increasing by one, it is beta decay.

Part (d) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (d) Step 2: Explanation

To complete the equation,

${}_{83}{}^{210}\mathrm{Bi}\to ?+{}_2{}^4\mathrm{He}$,

equate the mass number,

We get$210=?+4$,

So, the mass number is$206$.

Now, equating the atomic number ,

We get$83=?+2$,

So, the atomic number is$81$.

So, the element is thallium.

Hence, the equation is ${}_{83}{}^{210}\mathrm{Bi}\to {}_{81}{}^{210}\mathrm{Tl}+{}_2{}^4\mathrm{He}$ and as the atomic number is decreasing by two, it is alpha decay.

Part (e) Step 1: Given Information

We are given an incomplete equation and we have to complete it and also describe its type.

Part (e) Step 2: Explanation

To complete the equation,

$?\to {}_{39}{}^{89}\mathrm{Y}+{}_1{}^0\mathrm{e}$,

equate the mass number,

We get$?=89+0$,

So, the mass number is$89$.

Now, equating the atomic number ,

We get$?=39+1$,

So, the atomic number is$40$.

So, the element is zirconium.

Hence, the equation isdata-custom-editor="chemistry" ${}_{40}{}^{89}\mathrm{Zr}\to {}_{39}{}^{89}\mathrm{Y}+{}_1{}^0\mathrm{e}$ and as the atomic number is decreasing by one, it is positron emission.