EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
Question
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Chapter 18, Problem 47AE

(a)

Interpretation Introduction

Interpretation:

 The balanced alpha emission 511B has to be written.

Concept Introduction:

Alpha-emission involves liberation of a 24He. The general equation for alpha emission process is written as follows:

   ZAYZ2AX+24He

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

(a)

Expert Solution
Check Mark

Explanation of Solution

The nuclear reaction alpha emission is as follows:

  511BZ2AX+24He        (1)

The nuclei that should come for X in equation (1) must have two atomic number more than 5 and mass number that can add up 4 to yield 11 as net mass number, so the nucleus must be 37Li as indicated below.

  511B37Li+24He

(b)

Interpretation Introduction

Interpretation:

The balanced beta emission 3888Sr has to be written.

Concept Introduction:

Beta-emission involves liberation of a beta-particle that is represented as 10e. The general equation for beta emission process is written as follows:

   ZAYZ+1AX+10e

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

(b)

Expert Solution
Check Mark

Explanation of Solution

The nuclear reaction for beta emission is as follows:

  1229MgZ+1AX+10e        (2)

There are 29 mass number and 12 atomic number on left side of equation (2) and thus to get same net mass numbers and atomic number the nuclei X must possess 11 as atomic number and 29 as mass number. Hence the nuclei that should come is 3988Y.

  3888Sr3988Y+10e

(c)

Interpretation Introduction

Interpretation:

 The balanced neutron absorption 47107Ag has to be written.

Concept Introduction:

Neutron absorption involves capture of neutron that is represented as 01n. The general equation neutron absorption process is written as follows:

  ZAY+01nZA+1X

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

(c)

Expert Solution
Check Mark

Explanation of Solution

The nuclear reaction for neutron absorption is as follows:

  47107Ag+01nZA+1X        (3)

The nuclei that should come for X in equation (3) must have same atomic number and one more than mass number 107, so the nucleus X must be 47108Ag as indicated below.

  47107Ag+01n47108Ag

(d)

Interpretation Introduction

Interpretation:

The balanced proton emission 3941K has to be written.

Concept Introduction:

Proton-emission involves liberation of a 11p. The general equation for proton emission process is written as follows:

  ZAYZ1A1X+11p

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

Initial nuclide is denoted as Y.

Product nuclide is Y.

(d)

Expert Solution
Check Mark

Explanation of Solution

The nuclear reaction of proton emission is as follows:

  1941KZ1A1X+11p        (4)

The nuclei that should come for X in equation (4) must have one atomic number less than 39 and one mass number less than 41, so the nucleus must be 1840Ar as indicated below.

  1941K1840Ar+11p

(e)

Interpretation Introduction

Interpretation:

 The balanced for electron absorption of 51116Sb has to be written.

Concept Introduction:

Electron absorption involves capture of electron that is represented as 10e. The general equation electron absorption process is written as follows:

  ZAY+10eZ1AX

Here,

Mass number is denoted by A.

Atomic number denoted by Z.

The initial nuclide is denoted as Y.

The product nuclide is Y.

(e)

Expert Solution
Check Mark

Explanation of Solution

The nuclear reaction electron neutron absorption is as follows:

  51116Sb+10eZ1AX        (5)

The nuclei that should come for X in equation (5) must have same atomic number and on more than mass number 107, so the nucleus X must be 47108Ag as indicated below.

  51116Sb+10e50116Sn

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Chapter 18 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 18 - Prob. 5RQCh. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - Prob. 10RQCh. 18 - Prob. 11RQCh. 18 - Prob. 12RQCh. 18 - Prob. 13RQCh. 18 - Prob. 14RQCh. 18 - Prob. 15RQCh. 18 - Prob. 16RQCh. 18 - Prob. 17RQCh. 18 - Prob. 18RQCh. 18 - Prob. 19RQCh. 18 - Prob. 20RQCh. 18 - Prob. 21RQCh. 18 - Prob. 22RQCh. 18 - Prob. 23RQCh. 18 - Prob. 24RQCh. 18 - Prob. 25RQCh. 18 - Prob. 26RQCh. 18 - Prob. 27RQCh. 18 - Prob. 28RQCh. 18 - Prob. 29RQCh. 18 - Prob. 30RQCh. 18 - Prob. 31RQCh. 18 - Prob. 32RQCh. 18 - Prob. 33RQCh. 18 - Prob. 1PECh. 18 - Prob. 2PECh. 18 - Prob. 3PECh. 18 - Prob. 4PECh. 18 - Prob. 5PECh. 18 - Prob. 6PECh. 18 - Prob. 7PECh. 18 - Prob. 8PECh. 18 - Prob. 9PECh. 18 - Prob. 10PECh. 18 - Prob. 11PECh. 18 - Prob. 12PECh. 18 - Prob. 13PECh. 18 - Prob. 14PECh. 18 - Prob. 15PECh. 18 - Prob. 16PECh. 18 - Prob. 17PECh. 18 - Prob. 18PECh. 18 - Prob. 21AECh. 18 - Prob. 22AECh. 18 - Prob. 23AECh. 18 - Prob. 24AECh. 18 - Prob. 25AECh. 18 - Prob. 26AECh. 18 - Prob. 27AECh. 18 - Prob. 28AECh. 18 - Prob. 29AECh. 18 - Prob. 30AECh. 18 - Prob. 31AECh. 18 - Prob. 32AECh. 18 - Prob. 33AECh. 18 - Prob. 34AECh. 18 - Prob. 35AECh. 18 - Prob. 36AECh. 18 - Prob. 37AECh. 18 - Prob. 38AECh. 18 - Prob. 39AECh. 18 - Prob. 40AECh. 18 - Prob. 41AECh. 18 - Prob. 42AECh. 18 - Prob. 43AECh. 18 - Prob. 44AECh. 18 - Prob. 45AECh. 18 - Prob. 46AECh. 18 - Prob. 47AECh. 18 - Prob. 48AECh. 18 - Prob. 49AECh. 18 - Prob. 50AECh. 18 - Prob. 51AECh. 18 - Prob. 52AECh. 18 - Prob. 53AECh. 18 - Prob. 54CECh. 18 - Prob. 55CE
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