Chapter 19, Problem 2SP

(a)

To determine

The atomic numbers of $\text{Th}$ , $\text{Ra}$ , $\text{Ac}$ , $\text{Rn}$ , $\text{Po}$ , $\text{Pb}$ and $\text{Bi}$ using periodic table.

Answer to Problem 2SP

The atomic number of $\text{Th}$ is $90$ , atomic number of $\text{Ra}$ is 88, that of $\text{Ac}$ is 89, that of $\text{Rn}$ is $86$ , that of $\text{Po}$ is 84 , that of $\text{Pb}$ is 82 and that of $\text{Bi}$ is $83$.

Explanation of Solution

Given info: The decay reaction given in the question is $\text{Th}\Rightarrow \text{Ra}\Rightarrow \text{Ac}\Rightarrow \text{Th}\Rightarrow \text{Ra}\Rightarrow \text{Rn}\Rightarrow \text{Po}\Rightarrow \text{Pb}\Rightarrow \text{Bi}\Rightarrow \text{Po}\Rightarrow \text{Pb}$.

The elements present in reaction are Thorium, Radium, Actinium, Radon, Polonium, Lead, and Bismuth.

From the periodic table , the atomic number of Thorium is $90$ .

The atomic number of Radium is $88$ .

The atomic number of Actinium is $89$ .

The atomic number of Radon is $86$.

The atomic number of Polonium is $84$.

The atomic number of Lead is $82$.

The atomic number of Bismuth is $83$

Conclusion:

Thus, The atomic number of Thorium($\text{Th}$) is $90$ , atomic number of Radium($\text{Ra}$) is 88, that of Actinium($\text{Ac}$) is 89, that of Radon($\text{Rn}$) is $86$ , that of Polonium($\text{Po}$) is 84 , that of Lead($\text{Pb}$) is 82 and that of($\text{Bi}$) is $83$.

(b)

To determine

The reaction in the series that involved alpha decay and beta decay.

Answer to Problem 2SP

The reactions that involved alpha decay are $\text{Th}\Rightarrow \text{Ra}$, $\text{Ra}\Rightarrow \text{Rn}$ , $\text{Rn}\Rightarrow \text{Po}$ ,$\text{Po}\Rightarrow \text{Pb}$ and the reactions that involved beta decay are $\text{Rn}\Rightarrow \text{Ac}$ , $\text{Ac}\Rightarrow \text{Th}$, $\text{Pb}\Rightarrow \text{Bi}$ , $\text{Bi}\Rightarrow \text{Po}$.

Explanation of Solution

The alpha decay is a decay reaction in which unstable elements decay by emitting alpha particle (nucleus of Helium-4) and beta decay is the decay reaction in which parent nucleus emit an electron or positron to become stable.

Write the symbolic representation of the series reaction.

${}_{90}\text{Th}\Rightarrow {}_{88}\text{Ra}\Rightarrow {}_{89}\text{Ac}\Rightarrow {}_{90}\text{Th}\Rightarrow {}_{88}\text{Ra}\Rightarrow {}_{86}\text{Rn}\Rightarrow {}_{84}\text{Po}\Rightarrow {}_{82}\text{Pb}\Rightarrow {}_{83}\text{Bi}\Rightarrow {}_{84}\text{Po}\Rightarrow {}_{82}\text{Pb}$

In an alpha decay the atomic number of parent nucleus decreases by two and mass number of parent nucleus decreases by four.

By checking the change in the atomic number of the nucleus, can be differentiate alpha decay and beta decay.

The reactions that involved alpha decay are therefore, ${}_{90}\text{Th}\Rightarrow {}_{88}\text{Ra}$,${}_{90}\text{Th}\Rightarrow {}_{88}\text{Ra}$, ${}_{88}\text{Ra}\Rightarrow {}_{86}\text{Rn}$,${}_{86}\text{Rn}\Rightarrow {}_{84}\text{Po}$,${}_{84}\text{Po}\Rightarrow {}_{82}\text{Pb}$, ${}_{84}\text{Po}\Rightarrow {}_{82}\text{Pb}$.

In beta decay the atomic number of parent nucleus increases or decreases by one depending on the electron or positron emitted during the reaction and there is no change in the mass number.

The reactions that involved beta decay are therefore, ${}_{88}\text{Ra}\Rightarrow {}_{89}\text{Ac}$,${}_{89}\text{Ac}\Rightarrow {}_{90}\text{Th}$ , ${}_{82}\text{Pb}\Rightarrow {}_{83}\text{Bi}$,${}_{83}\text{Bi}\Rightarrow {}_{84}\text{Po}$.

Conclusion:

Thus, the reactions that involved alpha decay are $\text{Th}\Rightarrow \text{Ra}$, $\text{Ra}\Rightarrow \text{Rn}$ , $\text{Rn}\Rightarrow \text{Po}$ ,$\text{Po}\Rightarrow \text{Pb}$ and the reactions that involved beta decay are $\text{Rn}\Rightarrow \text{Ac}$ , $\text{Ac}\Rightarrow \text{Th}$, $\text{Pb}\Rightarrow \text{Bi}$ , $\text{Bi}\Rightarrow \text{Po}$.

(c)

To determine

The reaction equations for the first three decays in this series.

Answer to Problem 2SP

The reaction equation for the first three series are,

${}_{90}{\text{Th}}^{232}{\to }_{88}{\text{Ra}}^{228}+{}_2{\text{He}}^4$

${}_{88}{\text{Ra}}^{228}\to {}_{89}{\text{Ac}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

${}_{89}{\text{Ac}}^{228}\to {}_{90}{\text{Th}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

Explanation of Solution

A reaction equation gives complete idea about the fragments involved in a reaction and is should conserve the neutron number and proton number.

Write the symbolic representation of first three reaction.

${}_{90}{\text{Th}}^{232}\Rightarrow {}_{88}{\text{Ra}}^{228}\Rightarrow {}_{89}{\text{Ac}}^{228}{\Rightarrow }_{90}{\text{Th}}^{228}$

Here,

${}_{90}{\text{Th}}^{232}$ is thorium-232 nucleus

${}_{88}{\text{Ra}}^{228}$ is Radium-228 nucleus

${}_{89}{\text{Ac}}^{228}$ is actinium-228 nucleus

${}_{90}{\text{Th}}^{228}$ is thorium-228 nucleus

Since the reaction ${}_{90}{\text{Th}}^{232}{\Rightarrow }_{88}{\text{Ra}}^{228}$ is a alpha decay, it involves the emission of alpha particle (${}_2{\text{He}}^4$) and therefore the mass number of thorium decrease by four.

Write the reaction equation.

${}_{90}{\text{Th}}^{232}{\to }_{88}{\text{Ra}}^{228}+{}_2{\text{He}}^4$

Since the reaction ${}_{88}{\text{Ra}}^{228}\Rightarrow {}_{89}{\text{Ac}}^{228}$ is a beta decay, it involves the emission of beta particle. The increase in the atomic number indicate that it involves a beta minus decay. In a beta decay the mass number Ra does not changes and in beta decay a charge less, mass less particle emits. The name of the particle is neutrino.

Write the reaction equation.

${}_{88}{\text{Ra}}^{228}\to {}_{89}{\text{Ac}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

Since the reaction ${}_{89}{\text{Ac}}^{228}{\Rightarrow }_{90}{\text{Th}}^{228}$ is a beta decay, it involves the emission of beta particle. The increase in the atomic number indicate that it involves a beta minus decay. In the beta decay the mass number Ac does not changes and in beta decay a charge less, mass less particle emits. The name of the particle is neutrino.

Write the reaction equation.

${}_{88}{\text{Ra}}^{228}\to {}_{89}{\text{Ac}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

Conclusion:

Thus, the reaction equation for the first three series are,

${}_{90}{\text{Th}}^{232}{\to }_{88}{\text{Ra}}^{228}+{}_2{\text{He}}^4$

${}_{88}{\text{Ra}}^{228}\to {}_{89}{\text{Ac}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

${}_{89}{\text{Ac}}^{228}\to {}_{90}{\text{Th}}^{228}+{}_{-1}{\text{e}}^0+{}_0{\nu }^0$

(d)

To determine

The mass numbers for all of the isotopes in the series.

Answer to Problem 2SP

The mass number of all the isotopes are given in the top of the name in eth following reaction series.

$\begin{array}{l}{}_{90}{\text{Th}}^{232}\Rightarrow {}_{88}{\text{Ra}}^{228}\Rightarrow {}_{89}{\text{Ac}}^{228}{\Rightarrow }_{90}{\text{Th}}^{228}\Rightarrow {}_{88}{\text{Ra}}^{224}\Rightarrow {}_{86}{\text{Rn}}^{220}\Rightarrow {}_{84}{\text{Po}}^{216}\Rightarrow {}_{82}{\text{Pb}}^{212}\\ \Rightarrow {}_{83}{\text{Bi}}^{212}\Rightarrow {}_{84}{\text{Po}}^{212}\Rightarrow {}_{82}{\text{Pb}}^{208}\end{array}$

Explanation of Solution

The alpha decay is a decay reaction in which unstable elements decay by emitting alpha particle (nucleus of Helium-4) and beta decay is the decay reaction in which parent nucleus emit an electron or positron to become stable.

In beta decay the atomic number of parent nucleus increases or decreases by one depending on the electron or positron emitted during the reaction and there is no change in the mass number.

By considering above facts write the reaction series including the mass number of each isotopes.

$\begin{array}{l}{}_{90}{\text{Th}}^{232}\Rightarrow {}_{88}{\text{Ra}}^{228}\Rightarrow {}_{89}{\text{Ac}}^{228}{\Rightarrow }_{90}{\text{Th}}^{228}\Rightarrow {}_{88}{\text{Ra}}^{224}\Rightarrow {}_{86}{\text{Rn}}^{220}\Rightarrow {}_{84}{\text{Po}}^{216}\Rightarrow {}_{82}{\text{Pb}}^{212}\\ \Rightarrow {}_{83}{\text{Bi}}^{212}\Rightarrow {}_{84}{\text{Po}}^{212}\Rightarrow {}_{82}{\text{Pb}}^{208}\end{array}$

Thus, the mass number of all the isotopes are given in the top of the name in eth following reaction series.

$\begin{array}{l}{}_{90}{\text{Th}}^{232}\Rightarrow {}_{88}{\text{Ra}}^{228}\Rightarrow {}_{89}{\text{Ac}}^{228}{\Rightarrow }_{90}{\text{Th}}^{228}\Rightarrow {}_{88}{\text{Ra}}^{224}\Rightarrow {}_{86}{\text{Rn}}^{220}\Rightarrow {}_{84}{\text{Po}}^{216}\Rightarrow {}_{82}{\text{Pb}}^{212}\\ \Rightarrow {}_{83}{\text{Bi}}^{212}\Rightarrow {}_{84}{\text{Po}}^{212}\Rightarrow {}_{82}{\text{Pb}}^{208}\end{array}$