Chapter 18, Problem 76QRT

(a)

Interpretation Introduction

Interpretation:

Formation of each isotope in the given conversion has to be written in terms of balanced equation.

Explanation of Solution

Given reaction is that element 117 is synthesized by collision of berkelium-249 and calcium-48 nuclei.  The products formed are two isotopes of element 117.

One isotope is formed by emission of three neutrons.  As three neutrons are emitted, the mass number is reduced by three.  The balanced nuclear equation can be given as,

    $\text{C}\text{a}\text{+}\text{B}\text{k}\stackrel{}{\to }\text{U}\text{us}+3\text{n}$

Another isotope is formed by emission of four neutrons.  As four neutrons are emitted, the mass number is reduced by four.  The balanced nuclear equation can be given as,

    $\text{C}\text{a}\text{+}\text{B}\text{k}\stackrel{}{\to }\text{U}\text{us}+4\text{n}$

(b)

Interpretation Introduction

Interpretation:

Series of balanced nuclear equation has to be written when the lighter isotope of element 117 undergoes three successive alpha emissions has to be written.

Explanation of Solution

Lighter isotope is $\text{U}\text{us}$.

First Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 115 and mass number of 289.  The element with atomic number 115 is represented as $\text{U}\text{up}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{us}\stackrel{}{\to }\text{U}\text{up}+\text{H}\text{e}$

Second Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 113 and mass number of 285.  The element with atomic number 113 is represented as $\text{U}\text{ut}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{up}\stackrel{}{\to }\text{U}\text{ut}+\text{H}\text{e}$

Third Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 111 and mass number of 281.  The element with atomic number 111 is represented as $\text{R}\text{g}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{ut}\stackrel{}{\to }\text{R}\text{g}+\text{H}\text{e}$

(c)

Interpretation Introduction

Interpretation:

Series of balanced nuclear equation has to be written when the heavier isotope of element 117 undergoes three successive alpha emissions has to be written.

Explanation of Solution

Heavier isotope is $\text{U}\text{us}$.

First Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 115 and mass number of 290.  The element with atomic number 115 is represented as $\text{U}\text{up}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{us}\stackrel{}{\to }\text{U}\text{up}+\text{H}\text{e}$

Second Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 113 and mass number of 286.  The element with atomic number 113 is represented as $\text{U}\text{ut}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{up}\stackrel{}{\to }\text{U}\text{ut}+\text{H}\text{e}$

Third Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 111 and mass number of 282.  The element with atomic number 111 is represented as $\text{R}\text{g}$.  The balanced nuclear equation can be given as,

    $\text{U}\text{ut}\stackrel{}{\to }\text{R}\text{g}+\text{H}\text{e}$

Fourth Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 109 and mass number of 278.  The element with atomic number 109 is represented as $\text{M}\text{t}$.  The balanced nuclear equation can be given as,

    $\text{R}\text{g}\stackrel{}{\to }\text{M}\text{t}+\text{H}\text{e}$

Fifth Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 107 and mass number of 274.  The element with atomic number 107 is represented as $\text{B}\text{h}$.  The balanced nuclear equation can be given as,

    $\text{M}\text{t}\stackrel{}{\to }\text{B}\text{h}+\text{H}\text{e}$

Sixth Alpha emission:

When alpha particle emission takes place, the mass number is reduced by 4 and atomic number is reduced by 2.  The formed nucleus will have atomic number of 105 and mass number of 270.  The element with atomic number 105 is represented as $\text{D}\text{b}$.  The balanced nuclear equation can be given as,

    $\text{B}\text{h}\stackrel{}{\to }\text{D}\text{b}+\text{H}\text{e}$

(d)

Interpretation Introduction

Interpretation:

The Group the element 117 is present in the periodic table has to be identified.

Explanation of Solution

On looking into the periodic table, Element 117 is present in Group VIIA.  Element 117 is present below Astatine.

(e)

Interpretation Introduction

Interpretation:

Name of the Group where the element 117 is present has to be given.

Explanation of Solution

On looking into the periodic table, Element 117 is present in Group VIIA.  Group VIIA is known as halogen group.