Chapter 20, Problem 38QP

Interpretation Introduction

Interpretation:

The balanced nuclear equations for the given reactions are to be written and the component X is to be identified.

Concept introduction:

In a balanced radioactive decay

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

In the balanced radioactive decay, the subscript present on the individual particles represents the atomic number, while the superscript represents the atomic mass.

While writing the abbreviated form of any radioactive reaction, the reactant is written first followed by parenthesis, which contains the bombarded particle first and then the emitted particle. After this the product is written.

Answer to Problem 38QP

Solution:

(a) ${{}^{\text{15}}}_{\text{7}}{\text{N +}}_{\text{1}}{}^{\text{1}}\text{p }\to {}^{\text{12}}{}_{\text{6}}{\text{C + }}^{\text{4}}{}_{\text{2}}\alpha$ and X is $\text{N}$.

(b) ${{}^{\text{27}}}_{\text{13}}{\text{Al +}}_{\text{1}}{}^2\text{d }\to {}^{\text{25}}{}_{\text{12}}{\text{Mg + }}^{\text{4}}{}_{\text{2}}\alpha$ and X is ${{}^{\text{25}}}_{\text{12}}\text{Mg}$.

(c) ${{}^{\text{55}}}_{\text{25}}{\text{Mn +}}_0{}^1\text{n }\to {}^{56}{}_{25}\text{Mn + }\gamma$ and X is ${{}^{56}}_{25}\text{Mn}$.

Explanation of Solution

a) $\text{X(p,}\alpha {\text{)}}^{\text{12}}{}_{\text{6}}\text{C}$.

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is $\text{X}$.

Then, it is followed by the parentheses where the bombarded particle is written first, followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is ${{}^1}_{1}p$ while the emitted particle is ${{}^4}_2\alpha$.

At the end, the parenthesis is closed and the product is written.

Here, the product is ${{}^{12}}_6\text{C}$.

Thus, the equation can be written as

${\text{X +}}_{\text{1}}{}^{\text{1}}\text{p }\to {}^{\text{12}}{}_{\text{6}}{\text{C + }}^{\text{4}}{}_{\text{2}}\alpha$

Now, apply the balancing rules as

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

On the left side of the reaction, the sum of the atomic numbers is given by the expression as follows: $\text{1 + atomic number of X}$

On the right-hand side of the reaction, the sum of the atomic numbers is calculated as follows:

$6+\text{ 2 = 8}$

Thus, the atomic number of X is calculated as follows:

$(8)-(1\text{)}=7$

Similarly,

On the left side of the reaction, the sum of the mass numbers is given by the expression as follows:

$1\text{ + mass number of X}$

On the right-hand side of the reaction, the sum of the mass numbers is as follows:

$12\text{ + 4 = 16}$

Thus, the mass number of X is calculated as follows:

$(16)-(1\text{)}=15$

So, X will be ${{}^{15}}_7\text{N}$.

Hence, the balanced reaction is given as follows:

${{}^{\text{15}}}_{\text{7}}{\text{N +}}_{\text{1}}{}^{\text{1}}\text{p }\to {}^{\text{12}}{}_{\text{6}}{\text{C + }}^{\text{4}}{}_{\text{2}}\alpha$

b) ${{}^{\text{27}}}_{\text{13}}\text{Al(d,}\alpha \text{)X}$

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is ${{}^{\text{27}}}_{\text{13}}\text{Al}$

Then, it is followed by the parenthesis where the bombarded particle is written first followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is ${{}^2}_1\text{d}$ while the emitted particle is ${{}^4}_2\alpha$.

At the end, the parenthesis is closed and the product is written.

Here, the product is $\text{X}$.

Thus, the equation can be written as follows:

${{}^{\text{27}}}_{\text{13}}{\text{Al +}}_{\text{1}}{}^2\text{d }\to {\text{ X + }}^{\text{4}}{}_{\text{2}}\alpha$

Now, apply the balancing rules as follows:

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

On the left side of the reaction, the sum of the atomic numbers is given as follows:

$\text{13 + 1 = 14}$

On the right hand side of the reaction, the sum of the atomic numbers is

$\text{atomic number of X}+\text{ 2 }$

Thus, the atomic number of X is calculated as follows:

$(14)-(2\text{)}=12$

Similarly,

On the left side of the reaction, the sum of the mass numbers is calculated as follows:

$\text{27 + 2 = 29}$

On the right hand side of the reaction, the sum of the mass numbers is

$\text{Atomic mass of X + 4}$

Thus, the mass number of X is calculated as follows:

$(29)-(4\text{)}=25$

So, X will be ${{}^{\text{25}}}_{\text{12}}\text{Mg}$.

Hence, the balanced reaction is given as follows:

${{}^{\text{27}}}_{\text{13}}{\text{Al +}}_{\text{1}}{}^2\text{d }\to {}^{\text{25}}{}_{\text{12}}{\text{Mg + }}^{\text{4}}{}_{\text{2}}\alpha$

c) ${{}^{55}}_{25}\text{Mn(n,}\gamma \text{)X}$

It is known that while writing the abbreviated form, the reactant is written first.

Here, the reactant is ${{}^{55}}_{25}\text{Mn}$

Then, it is followed by the parenthesis where the bombarded particle is written first followed by the emitted particles. These are separated by commas.

Here, the bombarded particle is ${{}^1}_0\text{n}$, while the emitted particle is $\gamma$.

At the end, the parenthesis is closed and the product is written.

Here, the product is $\text{X}$.

Thus, the equation can be written as follows:

${{}^{\text{55}}}_{\text{25}}{\text{Mn +}}_0{}^1\text{n }\to \text{ X + }\gamma$

Now, apply the balancing rules as follows:

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

On the left side of the reaction, the sum of the atomic numbers is given as follows:

$\text{25 + 0 = 25}$

On the right hand side of the reaction, the sum of the atomic numbers is

$\text{Atomic number of X}+\text{ 0}$

Thus, the atomic number of X is calculated as follows:

$(25)-(0\text{)}=25$

Similarly,

On the left side of the reaction, the sum of the mass numbers is calculated as follows:

$\text{55 + 1 = 56}$

On the right hand side of the reaction, the sum of the mass numbers is

$\text{Atomic mass of X + 0}$

Thus, the mass number of X is calculated as follows:

$(56)-(0\text{)}=56$

So, X will be ${{}^{56}}_{25}\text{Mn}$

Hence, the balanced reaction is given as follows:

${{}^{\text{55}}}_{\text{25}}{\text{Mn +}}_0{}^1\text{n }\to {}^{56}{}_{25}\text{Mn + }\gamma$