Chapter 18, Problem 13PE

(a)

Interpretation Introduction

Interpretation:

Nuclear equation below has to be completed.

  ${}_{29}^{66}\text{Cu}{\to }_{30}^{66}\text{Zn}+\underset{\_}{}$

Concept Introduction:

Alpha particle $\left(\text{α}\right)$ represents nucleus of helium atom. It contains two protons and two neutrons. Emission of this alpha particle is called alpha decay. Symbol of alpha particle is ${}_2^4\text{He}$.

General equation for alpha decay is as follows:

  ${}_Z^A\text{X}{\to }_{Z-2}^{A-4}\text{Y}{+}_2^4\text{He}$

Here,

$A$ is the mass or nucleon number.

$Z$ is the atomic number.

$\text{X}$ is the symbol of the element.

Beta emission $\left({\text{β}}^{-}\right)$ process represents conversion of neutron into 1 proton and 1 electron. This proton stays within nucleus and electron is emitted out of atom. This electron is called the beta particle $\left({}_{-1}^0\text{e}\right)$.

The generic equation of the beta decay is as follows:

  ${}_Z^A\text{X}{\to }_{Z+1}^A\text{Y}{+}_{-1}^0\text{e}$

Here,

$A$ is the mass or nucleon number.

$Z$ is the atomic number.

$\text{X}$ and Y is the symbol of the element.

Gamma rays $\left(\text{γ}\right)$ are energy photons with more energy than X-rays. Loss of gamma ray causes no change in mass numbers as well as atomic numbers.

Explanation of Solution

Skeleton equation for transition is as follows:

  ${}_{29}^{66}\text{Cu}{\to }_{30}^{66}\text{Zn}+\underset{\_}{}$

Since this transition causes atomic number to increase by 1 thus this transition is called beta emission. Resultant nuclear equation for beta emission from ${}_{29}^{66}\text{Cu}$ is as follows:

  ${}_{29}^{66}\text{Cu}{\to }_{30}^{66}\text{Zn}+\underset{\_}{{}_{-1}^0\text{e}}$

(b)

Interpretation Introduction

Interpretation:

Nuclear equation below has to be completed.

  ${}_{-1}^0\text{e}+\underset{\_}{}{\to }_3^7\text{Li}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Skeleton equation for transition is as follows:

  ${}_{-1}^0\text{e}+\underset{\_}{}{\to }_3^7\text{Li}$

Since this transition is electron capture thus it causes atomic number to decrease by 1. According to periodic table, element with atomic number 4 is beryllium $\left(\text{Be}\right)$. Resultant nuclear equation for electron capture is as follows:

  ${}_{-1}^0\text{e}+\underset{\_}{{}_4^7\text{Be}}{\to }_3^7\text{Li}$

(c)

Interpretation Introduction

Interpretation:

Nuclear equation below has to be completed.

  ${}_{13}^{27}\text{Al}{+}_2^4\text{He}{\to }_{14}^{30}\text{Si}+\underset{\_}{}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Skeleton equation for transition is as follows:

  ${}_{13}^{27}\text{Al}{+}_2^4\text{He}{\to }_{14}^{30}\text{Si}+\underset{\_}{}$

Since this transition is nuclear fission between ${}_{13}^{27}\text{Al}$ and ${}_2^4\text{He}$ thus sum of atomic number must be equal to sum of atomic number on right side. Also, sum of mass number on left side must be equal to sum of mass number on right side.

Sum of mass number on left side is 31 and on right side is 30. Thus missed particle must have 1 mass number.

Sum of atomic number on left side is 15 and on right side is 14. Thus missed particle must have 1 atomic number.

According to periodic table, element with atomic number 1 is hydrogen.

Hence resultant nuclear equation can be completed as follows:

  ${}_{13}^{27}\text{Al}{+}_2^4\text{He}{\to }_{14}^{30}\text{Si}+\underset{\_}{{}_1^1\text{H}}$

(d)

Interpretation Introduction

Interpretation:

Nuclear equation below has to be completed.

  ${}_{37}^{85}\text{Rb}+\underset{\_}{}{\to }_{35}^{82}\text{Br}{+}_2^4\text{He}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Skeleton equation for transition is as follows:

  ${}_{37}^{85}\text{Rb}+\underset{\_}{}{\to }_{35}^{82}\text{Br}{+}_2^4\text{He}$

Since this transition is nuclear fission thus sum of atomic number must be equal to sum of atomic number on right side. Also, sum of mass number on left side must be equal to sum of mass number on right side.

Sum of mass number on left side is 85 and on right side is 86. Thus missed particle must have 1 mass number.

Sum of atomic number on left side is 37 and on right side is 37. Thus missed particle must have 0 atomic number.

Atomic particle with 1 mass number and zero atomic number is ${}_0^1\text{n}$.

Hence resultant nuclear equation can be completed as follows:

  ${}_{37}^{85}\text{Rb}+\underset{\_}{{}_0^1\text{n}}{\to }_{35}^{82}\text{Br}{+}_2^4\text{He}$