Chapter 20, Problem 6QP

Interpretation Introduction

Interpretation:

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

Concept Introduction:

In a balanced radioactive decay:

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

And

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

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 6QP

Solution:

(a)

The balanced nuclear reaction is $(b)(c){}_{(f)53}^{(h)135}(j)\text{I }{\to }_{(m)54}^{(o)135}(q)\text{Xe}{+}_{(t)-1}^0(w)\text{β}.(x)$ (y)

(z)

The balanced nuclear reaction is $(aa)(bb){}_{(ee)19}^{(gg)40}(ii)\text{K }{\to }_{(ll)-1}^0(oo){(qq)\text{β+}}_{(ss)20}^{(uu)40}(ww)\text{Ca}(xx)$ (yy).

(zz)

The balanced nuclear reaction is $(aaa)(bbb){}_{(eee)27}^{(ggg)59}(iii){(kkk)\text{Co + }}^{\text{1}}(nnn){}_{\text{0}}(rrr)\text{n}{\to }_{(uuu)25}^{(www)56}(yyy)\text{Mn}{+}_2^4(dddd)\alpha (eeee)$ (ffff).

(gggg)

The balanced nuclear reaction is $(hhhh)(iiii){(kkkk){}^{(nnnn)235}(pppp)}_{(rrrr)\text{92}}(tttt){(vvvv)\text{U + }}^{\text{1}}(yyyy){}_{\text{0}}(ccccc)\text{n}\to {}_{(fffff)40}^{(hhhhh)99}(jjjjj)\text{Zr}{+}_{(mmmmm)52}^{(ooooo)135}(qqqqq){(sssss)\text{Te + 2}}^1(vvvvv){}_0(zzzzz)\text{n}.(aaaaaa)$ (bbbbbb)

Explanation of Solution

a) ${}_{53}^{135}\text{I }{\to }_{54}^{135}\text{Xe}+\text{X}\text{.}$

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 the atomic numbers is $53$.

On the right side of the reaction, the sum of the atomic numbers is $\left(54\right)\text{+ }\left(\text{atomic number of X}\right)$.

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

$(53)-(54\text{)}=-1$

Similarly,

On the left side of the reaction, the sum of the mass number is $135$.

On the right side of the reaction, the sum of the mass numbers is $\left(135\right)+\left(\text{mass number of X}\right)\text{.}$

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

$(135)-(135\text{)}=0$

So, X will be ${}_{-1}^0\text{β}.$

The balanced nuclear reaction is as follows:

${}_{53}^{135}\text{I }{\to }_{54}^{135}\text{Xe}{+}_{-1}^0\text{β}.$

b) ${}_{19}^{40}\text{K }{\to }_{-1}^0\text{β+X}$

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 the atomic numbers is $19$.

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

Thus, the atomic number of X is as follows:

$(19)+(1\text{)}=20$

Similarly,

On the left side of the reaction, the sum of the mass numbers is $40$.

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

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

$\left(40\right)-\left(0\right)=40$

So, X will be ${}_{20}^{40}\text{Ca}$.

The balanced nuclear reaction is as follows:

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

c) ${}_{27}^{59}{\text{Co + }}^{\text{1}}{}_{\text{0}}\text{n}{\to }_{25}^{56}\text{Mn}+\text{X}$

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 the atomic numbers is as follows:

$\left(27\right)+\left(0\right)=27$

On the right side of the reaction, the sum of the atomic numbers is $\text{25+ (atomic number of X)}$.

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

$\left(27\right)-\left(25\right)=2$

Similarly,

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

$\left(59\right)+\left(1\right)=60$

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

Thus, the mass number of X is as follows:

$\left(60\right)-\left(56\right)=4$

So, X will be ${}_2^4\alpha$.

The balanced nuclear reaction is as follows:

${}_{27}^{59}{\text{Co + }}^{\text{1}}{}_{\text{0}}\text{n}{\to }_{25}^{56}\text{Mn}{+}_2^4\alpha$

d) ${{}^{235}}_{\text{92}}{\text{U + }}^{\text{1}}{}_{\text{0}}\text{n}\to {}_{40}^{99}\text{Zr}{+}_{52}^{135}\text{Te + 2X}$

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 as follows:

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

On the right side of the reaction, the sum of the atomic numbers is $\left(\text{40}\right)\text{+}\left(\text{52}\right)\text{+ 2}\times \text{(atomic number of X)}$.

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

$\left(\frac{(92)-(\text{40+52)}}{2}\right)=0$

Similarly,

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

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

On the right side of the reaction, the sum of the mass numbers is $\left(\text{99}\right)\text{+ }\left(\text{135}\right)\text{+ 2}\times \text{(mass number of X)}$.

Thus, the mass number of X is as follows:

$\frac{(236)-(\text{99 + 135)}}{2}=1$

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

Thus, the balanced nuclear reaction is as follows:

${{}^{235}}_{\text{92}}{\text{U + }}^{\text{1}}{}_{\text{0}}\text{n}\to {}_{40}^{99}\text{Zr}{+}_{52}^{135}{\text{Te + 2}}^1{}_0\text{n}.$