Chapter 20, Problem 5QP

Interpretation Introduction

Interpretation:

The unknown atom, X, is to be identified in the given nuclear reactions.

Concept Introduction:

In a balanced radioactive decay:

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

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

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

Answer to Problem 5QP

Solution:

(a)

The term X is $(b)(c){}_{(f)11}^{(h)23}(j)\text{Na}\text{.}(k)$ (l)

(m)

The term X is $(n)(o){}_1^1(t)\text{H}.(u)$ (v)

(w)

The term X is $(x)(y){(aa){}^1(ee)}_0(hh)\text{n}(ii)$ (jj).

(kk)

The term X is $(ll)(mm){}_{(pp)26}^{(rr)56}(tt)\text{Fe}(uu)$ (vv).

(ww)

The term X is $(xx)(yy){}_{(bbb)-1}^0(eee)\text{β}.(fff)$ (ggg)

Explanation of Solution

a) ${}_{12}^{26}{\text{Mg +}}^{\text{1}}{}_1\text{p }\to \text{α}+\text{X}\text{.}$

The atomic mass and atomic number of $\alpha$ are $4\text{ and 2 }$, respectively.

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 atomic numbers is calculated as follows:

$\left(\text{12}\right)\text{ + }\left(\text{1}\right)=\text{13}$

On the right side of the reaction, the sum of atomic numbers is $\left(2\right)\text{+ }\left(\text{atomic number of X}\right)$.

Thus, the atomic number of X is as follows:

$(13)-(2\text{)}=11.$

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

$\left(\text{26}\right)+\left(\text{1}\right)\text{ = 27}$

On the right side of the reaction, the sum of mass numbers is $\left(\text{4}\right)+\left(\text{mass number of X}\right)\text{.}$

Thus, the mass number of X is as follows:

$(27)-(4\text{)}=23$

So, X will be ${}_{11}^{23}\text{Na}\text{.}$

b) ${}_{27}^{59}{\text{Co+}}_1^2\text{H}{\to }_{27}^{60}\text{Co + X}$

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 atomic numbers is calculated as follows:

$\left(27\right)+\left(1\right)=28$

On the right side of the reaction, the sum of atomic numbers is $\left(\text{27}\right)\text{ + }\left(\text{atomic number of X}\right)$.

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

$(28)-(27)=1$

Similarly,

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

$\left(59\right)+\left(2\right)=61$

On the right side of the reaction, the sum of mass numbers is $\left(60\right)+\left(\text{mass number of X}\right)$.

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

$(61)-(60)=1$

So, X will be ${}_1^1\text{H}$.

c) ${{}^{235}}_{\text{92}}{\text{U + }}^{\text{1}}{}_{\text{0}}\text{n}\to {}_{36}^{94}{\text{Kr+}}_{56}^{139}\text{Ba + 3X}$

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 atomic numbers is as follows:

$\left(92\right)\text{ + }\left(\text{0}\right)\text{ = 92}$

On the right side of the reaction, the sum of atomic numbers is $\text{36 + 56+ 3}\times \text{(atomic number of X)}$.

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

$\frac{(92)-(36+56)}{3}=0$

Similarly,

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

$\left(235+1\right)=236$

On the right side of the reaction, the sum of mass numbers is $\text{94 + 139 + 3}\times \text{(mass number of X)}$.

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

$\left(\frac{(236)-(94+139)}{3}\right)=1$

So, X will be ${{}^1}_0\text{n}$.

d) ${}_{24}^{53}{\text{Cr +}}_2^4\alpha {\to }_0^1\text{n}+\text{X}\text{.}$

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 atomic numbers is calculated as follows:

$\left(24\right)+\left(2\right)=26$

On the right side of the reaction, the sum of atomic numbers is $\left(0\right)+\text{(atomic number of X)}$.

Thus, the atomic number of X is as follows:

$\left(26\right)-\left(0\right)=26$

Similarly,

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

$\left(53\right)+\left(4\right)=57$

On the right side of the reaction, the sum of mass numbers is $\left(1\right)+\text{(mass number of X)}$.

Thus, the mass number of X is as follows:

$\left(57\right)-\left(1\right)=56$

So, X will be ${}_{26}^{56}\text{Fe}$.

e) ${}_8^{20}\text{O }{\to }_9^{20}\text{F}+\text{X}\text{.}$

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 atomic numbers is $8$.

On the right side of the reaction, the sum of atomic numbers is $\left(9\right)+\text{(atomic number of X)}$.

Thus, atomic number of X is calculated as follows:

$\left(8\right)-\left(9\right)=-1$

Similarly,

On the left side of the reaction, the sum of mass numbers is $20$.

On the right side of the reaction, the sum of mass numbers is $\left(20\right)+\text{(mass number of X)}$.

Thus, the mass number of X is as follows:

$\left(20\right)-\left(20\right)=0$

So, X will be ${}_{-1}^0\text{β}$.