EBK CHEMISTRY: ATOMS FIRST
EBK CHEMISTRY: ATOMS FIRST
3rd Edition
ISBN: 8220103675505
Author: Burdge
Publisher: YUZU
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Chapter 20, Problem 20.94QP

(a)

Interpretation Introduction

Interpretation:

Suitable assumption should be drawn for the given calculation.

Concept Introduction:

  • Half-life period: The time required to reduce half of its original value.
  • This is denoted by: t1/2
  • This term is popularly use in nuclear chemistry, to describe the stability & instability of atoms or radioactive decay.
  • For determination of first order rate constant K in half –life period we use

    K=0.693t1/2 Equation.

  • The half-life in first order does not depend upon the initial concentration.

Formula:

The number of disintegration is directly proportional to the number of radioactive nuclei

lndecay rate of old sampledecay rate of fresh sample=-kt

The activity is given by:

activity=numberdecayUnit time=kN

Age calculation:

lnNtN0=-kt

Calculate the age and rate constant:

K=0.693t1/2

Given:

The half-life of Mg27 is 9.50 min.

Initially 4.20 ×1012years Mg27 nuclei present.

To determine: Determine the remaining Mg27 nuclei after 30.00 min

(b)

Interpretation Introduction

Interpretation:

Suitable assumption should be drawn for the given calculation.

Concept Introduction:

  • Half-life period: The time required to reduce half of its original value.
  • This is denoted by: t1/2
  • This term is popularly use in nuclear chemistry, to describe the stability & instability of atoms or radioactive decay.
  • For determination of first order rate constant K in half –life period we use

    K=0.693t1/2 Equation.

  • The half-life in first order does not depend upon the initial concentration.

Formula:

The number of disintegration is directly proportional to the number of radioactive nuclei

lndecay rate of old sampledecay rate of fresh sample=-kt

The activity is given by:

activity=numberdecayUnit time=kN

Age calculation:

lnNtN0=-kt

Calculate the age and rate constant:

K=0.693t1/2

Given:

The half-life of Mg27 is 9.50 min.

Initially 4.20 ×1012years Mg27 nuclei present.

To determine: The Mg27 activity (in Ci) at t=0 and t=30.0min is calculated.

(c)

Interpretation Introduction

Interpretation:

Suitable assumption should be drawn for the given calculation.

Concept Introduction:

  • Half-life period: The time required to reduce half of its original value.
  • This is denoted by: t1/2
  • This term is popularly use in nuclear chemistry, to describe the stability & instability of atoms or radioactive decay.
  • For determination of first order rate constant K in half –life period we use

    K=0.693t1/2 Equation.

  • The half-life in first order does not depend upon the initial concentration.

Formula:

The number of disintegration is directly proportional to the number of radioactive nuclei

lndecay rate of old sampledecay rate of fresh sample=-kt

The activity is given by:

activity=numberdecayUnit time=kN

Age calculation:

lnNtN0=-kt

Calculate the age and rate constant:

K=0.693t1/2

Given:

The half-life of Mg27 is 9.50 min.

Initially 4.20 ×1012years Mg27 nuclei present.

To determine: Probability of finding Mg27 nuclei decay during a 1 s interval

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Chapter 20 Solutions

EBK CHEMISTRY: ATOMS FIRST

Ch. 20.2 - Prob. 20.2.2SRCh. 20.2 - What is the change in mass (in ka) for the...Ch. 20.3 - Prob. 20.3WECh. 20.3 - Prob. 3PPACh. 20.3 - Prob. 3PPBCh. 20.3 - Prob. 20.4WECh. 20.3 - Prob. 4PPACh. 20.3 - Prob. 20.3.1SRCh. 20.3 - Prob. 20.3.2SRCh. 20.4 - Prob. 20.5WECh. 20.4 - Prob. 5PPACh. 20.4 - Prob. 5PPBCh. 20.4 - Prob. 5PPCCh. 20.4 - Prob. 20.4.1SRCh. 20.4 - Prob. 20.4.2SRCh. 20 - Prob. 20.1QPCh. 20 - Prob. 20.2QPCh. 20 - Prob. 20.3QPCh. 20 - Prob. 20.4QPCh. 20 - Prob. 20.5QPCh. 20 - Prob. 20.6QPCh. 20 - Prob. 20.7QPCh. 20 - Prob. 20.8QPCh. 20 - Prob. 20.9QPCh. 20 - Prob. 20.10QPCh. 20 - Prob. 20.11QPCh. 20 - Prob. 20.12QPCh. 20 - Prob. 20.13QPCh. 20 - Prob. 20.14QPCh. 20 - Prob. 20.15QPCh. 20 - Prob. 20.16QPCh. 20 - Prob. 20.17QPCh. 20 - Prob. 20.18QPCh. 20 - Prob. 20.19QPCh. 20 - Prob. 20.20QPCh. 20 - Prob. 20.21QPCh. 20 - Prob. 20.22QPCh. 20 - Prob. 20.23QPCh. 20 - Prob. 20.24QPCh. 20 - Prob. 20.25QPCh. 20 - Prob. 20.26QPCh. 20 - Prob. 20.27QPCh. 20 - Prob. 20.28QPCh. 20 - Prob. 20.29QPCh. 20 - Prob. 20.30QPCh. 20 - Prob. 20.31QPCh. 20 - Prob. 20.32QPCh. 20 - Prob. 20.33QPCh. 20 - Prob. 20.34QPCh. 20 - Prob. 20.35QPCh. 20 - Prob. 20.36QPCh. 20 - Prob. 20.37QPCh. 20 - Prob. 20.38QPCh. 20 - Prob. 20.39QPCh. 20 - Prob. 20.1VCCh. 20 - Prob. 20.3VCCh. 20 - Prob. 20.4VCCh. 20 - Prob. 20.40QPCh. 20 - Prob. 20.41QPCh. 20 - Prob. 20.42QPCh. 20 - Prob. 20.43QPCh. 20 - Prob. 20.44QPCh. 20 - Prob. 20.45QPCh. 20 - Prob. 20.46QPCh. 20 - Prob. 20.47QPCh. 20 - Prob. 20.48QPCh. 20 - Prob. 20.49QPCh. 20 - Prob. 20.50QPCh. 20 - Prob. 20.51QPCh. 20 - Prob. 20.52QPCh. 20 - Prob. 20.53QPCh. 20 - Prob. 20.54QPCh. 20 - Prob. 20.55QPCh. 20 - Prob. 20.56QPCh. 20 - Prob. 20.57QPCh. 20 - Prob. 20.58QPCh. 20 - Prob. 20.59QPCh. 20 - Prob. 20.60QPCh. 20 - Prob. 20.61QPCh. 20 - Prob. 20.62QPCh. 20 - Prob. 20.63QPCh. 20 - Prob. 20.64QPCh. 20 - Prob. 20.65QPCh. 20 - Prob. 20.66QPCh. 20 - Prob. 20.67QPCh. 20 - Prob. 20.68QPCh. 20 - Prob. 20.69QPCh. 20 - Prob. 20.70QPCh. 20 - Prob. 20.71QPCh. 20 - Prob. 20.72QPCh. 20 - Prob. 20.73QPCh. 20 - Prob. 20.74QPCh. 20 - Prob. 20.75QPCh. 20 - Prob. 20.76QPCh. 20 - Prob. 20.77QPCh. 20 - Prob. 20.78QPCh. 20 - Prob. 20.79QPCh. 20 - Prob. 20.80QPCh. 20 - Prob. 20.81QPCh. 20 - Prob. 20.82QPCh. 20 - Prob. 20.83QPCh. 20 - Prob. 20.84QPCh. 20 - Prob. 20.85QPCh. 20 - Prob. 20.86QPCh. 20 - Prob. 20.87QPCh. 20 - Prob. 20.88QPCh. 20 - Prob. 20.89QPCh. 20 - Prob. 20.90QPCh. 20 - Prob. 20.91QPCh. 20 - Prob. 20.92QPCh. 20 - Prob. 20.93QPCh. 20 - Prob. 20.94QPCh. 20 - Prob. 20.95QPCh. 20 - Prob. 20.96QPCh. 20 - Prob. 20.97QPCh. 20 - Prob. 20.98QPCh. 20 - Prob. 20.99QPCh. 20 - Prob. 20.100QPCh. 20 - Prob. 20.101QP
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