Chapter 24.2, Problem 8PP

Interpretation Introduction

(a)

Interpretation:

The balanced nuclear equation for decay of bismuth-212 needs to be written.

Concept introduction:

Nuclear reactions are expressed by balanced nuclear reactions just as chemical reactions are expressed by balanced nuclear reactions. However, in balanced chemical equations, numbers and type of atoms are conserved but in nuclear equations, mass numbers and charges are conserved.

Answer to Problem 8PP

Balanced nuclear equation for decay of bismuth-212 is,

$\begin{array}{l}\text{B}\text{i}\to \text{H}\text{e}\text{+}\text{T}\text{i}\\ \text{A}\end{array}$

$\begin{array}{l}\text{T}\text{i}\to e+\text{P}\text{b}\\ \text{B}\end{array}$

Explanation of Solution

The given diagram of decay of bismuth-212 is as follows:

From the given diagram it is clear that decay of bismuth-212 gives alpha particle and A. Alpha particle releases means the decay is alpha decay. In alpha decay, mass number decrease by 4 and atomic number decreases by 2. So, the nuclear equation is,

$\begin{array}{l}\text{B}\text{i}\to \text{H}\text{e}\text{+}\text{T}\text{i}\\ \text{A}\end{array}$

A again undergoes decay to form $\text{β}$ particle and B. In beta decay, atomic number increases by one and no change in the mass number. So, the nuclear equation is,

$\begin{array}{l}\text{T}\text{i}\to {\text{β}}^{+}+\text{P}\text{b}\\ \text{T}\text{i}\to e+\text{P}\text{b}\\ \text{B}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The isotopes A and B are to be identified in the decay of bismuth-212.

Concept introduction:

Isotopes are atoms of same element whose number of neutrons is different. Isotopes having unstable nuclei are termed as radioisotopes. These unstable nuclei emit radiation to attain more stable atomic configuration is termed as radioactive decay.

Answer to Problem 8PP

A is $\text{T}\text{i}$ and B is $\text{P}\text{b}$.

Explanation of Solution

The given diagram of decay of bismuth-212 is as follows:

From the given diagram it is clear that decay of bismuth-212 gives alpha particle and A. Alpha particle releases means the decay is alpha decay. In alpha decay, mass number decrease by 4 and atomic number decreases by 2. So, the nuclear equation is,

$\begin{array}{l}\text{B}\text{i}\to \text{H}\text{e}\text{+}\text{T}\text{i}\\ \text{A}\end{array}$

A again undergoes decay to form $\text{β}$ particle and B. In beta decay, atomic number increases by one and no change in the mass number. So, the nuclear equation is,

$\begin{array}{l}\text{T}\text{i}\to {\text{β}}^{+}+\text{P}\text{b}\\ \text{T}\text{i}\to e+\text{P}\text{b}\\ \text{B}\end{array}$

So, A is $\text{T}\text{i}$ and B is $\text{P}\text{b}$.