Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 24, Problem 24.94P

(a)

Interpretation Introduction

Interpretation:

Number of Rads (Dose) received by 85 kg worker for 16 h has to be determined using the given data.

Concept Introduction:

The half–life period for first order reaction is as follows:

  t1/2=ln 2kWhere,k = Rate constantt1/2= half-life

Decay rate can be calculated using the given equation as follows,

  Deacy rate =kNwhere, k = Deacy constantN = No. of radioactive nuclei

Mass of a substance can be determined from molar mass as given,

  Mass = No. of moles×Molar mass

Radiation Dose is the dose quantity that explains the effects of ionizing radiation. It describes the quantity of radiation energy that is absorbed by a matter. The units commonly used to measure dose are gray (Gy) and rad.

  1 Gy = 1 J/kg1 rad = 0.01 J/kg = 0.01 Gy

Unit conversion:

  1 MeV = (106 eV)(1.602×1019 J1 eV)

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

Mass of worker = 85 kgEnergy released in each disintegration = 5.15 MeVHalf-life time, t1/2 of 94239Pu = 2.41×104 yrMass of 94239Pu in the dust = 1.00 μg = 1.00×106 g

  • Number of 239Pu is determined as follows,

No.of atoms = Mass Molar mass×Avogadro's number= 1×106 g 239Pu239 g/mol239Pu×6.022×1023 atoms/mol 239Pu= 2.5196653×1015 atoms 239Pu

  • Rate constant, k is determined as shown,

  t1/2 = ln 2kk = (ln 22.41×104 yr)(1 yr365.25 d)(1 d24 h)(1 h3600 s)= 9.113904×1013 yr1

  • Decay constant can be calculated using the given equation as follows,

Deacy rate, A= kN = ( 9.111969×1013 yr1)(2.5196653×1015 atoms )(1 disintegration1 atom) = 2.2963988×103 dps

  • Number of Rads (Dose) received by 85 kg human is determined as given below,

Dose = (2.2963988×103 dps85 kg)(yr)(5.15 MeVdisintegration)(1.602×1013 J1 MeV)(1 rad0.01 J/kg)(3600 s1 h)(16 h)= 1.2838686×104 rad= 1.28×104 rad

Number of Rads (Dose) received by 85 kg worker receive for 16 h is 1.28×10-4 rad.

(b)

Interpretation Introduction

Interpretation:

Number of Rads (Dose) received by 65 kg worker for 16 h has to be determined using the given data.

Concept Introduction:

The half–life period for first order reaction is as follows:

  t1/2=ln 2kWhere,k = Rate constantt1/2= half-life

Decay rate can be calculated using the given equation as follows,

  Deacy rate =kNwhere, k = Deacy constantN = No. of radioactive nuclei

Mass of a substance can be determined from molar mass as given,

  Mass = No. of moles×Molar mass

Radiation Dose is the dose quantity that explains the effects of ionizing radiation. It describes the quantity of radiation energy that is absorbed by a matter. The units commonly used to measure dose are gray (Gy) and rad.

  1 Gy = 1 J/kg1 rad = 0.01 J/kg = 0.01 Gy

Unit conversion:

  1 MeV = (106 eV)(1.602×1019 J1 eV)

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

Mass of worker = 85 kgEnergy released in each disintegration = 5.15 MeVHalf-life time, t1/2 of 94239Pu = 2.41×104 yrMass of 94239Pu in the dust = 1.00 μg = 1.00×106 g

  • Number of 239Pu is determined as follows,

No.of atoms = Mass Molar mass×Avogadro's number= 1×106 g 239Pu239 g/mol239Pu×6.022×1023 atoms/mol 239Pu= 2.5196653×1015 atoms 239Pu

  • Rate constant, k is determined as shown,

  t1/2 = ln 2kk = (ln 22.41×104 yr)(1 yr365.25 d)(1 d24 h)(1 h3600 s)= 9.113904×1013 yr1

  • Decay constant can be calculated using the given equation as follows,

Deacy rate, A= kN = ( 9.111969×1013 yr1)(2.5196653×1015 atoms )(1 disintegration1 atom) = 2.2963988×103 dps

  • Number of Rads (Dose) received by 85 kg human is determined as given below,

  Dose = ((2.2963988×103 dps85 kg)(yr)(5.15 MeVdisintegration)(1.602×1013 J1 MeV)(1 rad0.01 J/kg)(3600 s1 h)(16 h))= 1.2838686×104 rad

Number of Rads (Dose) received by 85 kg worker receive for 16 h is 1.2838686×104 rad.

  • Convert Number of Rads to Gy,

  1 rad = 0.01 GyDose = (1.2838686×104 rad)(0.01 Gy/rad)= 1.2838686×106= 1.28×106 Gy

Number of Gy (Dose) received by 85 kg worker receive for 16 h is 1.28×10-6 Gy.

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Chapter 24 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 24.2 - Prob. 24.6AFPCh. 24.2 - Prob. 24.6BFPCh. 24.6 - Prob. 24.7AFPCh. 24.6 - Prob. 24.7BFPCh. 24.7 - Prob. B24.1PCh. 24.7 - Prob. B24.2PCh. 24.7 - Prob. B24.3PCh. 24.7 - Prob. B24.4PCh. 24 - Prob. 24.1PCh. 24 - Prob. 24.2PCh. 24 - Prob. 24.3PCh. 24 - Prob. 24.4PCh. 24 - Prob. 24.5PCh. 24 - Prob. 24.6PCh. 24 - Prob. 24.7PCh. 24 - Prob. 24.8PCh. 24 - Prob. 24.9PCh. 24 - Prob. 24.10PCh. 24 - Prob. 24.11PCh. 24 - Prob. 24.12PCh. 24 - Prob. 24.13PCh. 24 - Prob. 24.14PCh. 24 - Prob. 24.15PCh. 24 - Prob. 24.16PCh. 24 - Prob. 24.17PCh. 24 - Prob. 24.18PCh. 24 - Prob. 24.19PCh. 24 - Prob. 24.20PCh. 24 - Prob. 24.21PCh. 24 - Prob. 24.22PCh. 24 - Prob. 24.23PCh. 24 - Prob. 24.24PCh. 24 - Prob. 24.25PCh. 24 - Prob. 24.26PCh. 24 - Prob. 24.27PCh. 24 - Prob. 24.28PCh. 24 - Prob. 24.29PCh. 24 - Prob. 24.30PCh. 24 - Prob. 24.31PCh. 24 - Prob. 24.32PCh. 24 - Prob. 24.33PCh. 24 - Prob. 24.34PCh. 24 - Prob. 24.35PCh. 24 - Prob. 24.36PCh. 24 - Prob. 24.37PCh. 24 - Prob. 24.38PCh. 24 - Prob. 24.39PCh. 24 - Prob. 24.40PCh. 24 - Prob. 24.41PCh. 24 - Prob. 24.42PCh. 24 - Prob. 24.43PCh. 24 - Prob. 24.44PCh. 24 - Prob. 24.45PCh. 24 - Prob. 24.46PCh. 24 - Prob. 24.47PCh. 24 - Prob. 24.48PCh. 24 - Prob. 24.49PCh. 24 - Prob. 24.50PCh. 24 - Prob. 24.51PCh. 24 - Prob. 24.52PCh. 24 - Prob. 24.53PCh. 24 - Prob. 24.54PCh. 24 - Prob. 24.55PCh. 24 - Prob. 24.56PCh. 24 - Prob. 24.57PCh. 24 - Prob. 24.58PCh. 24 - Prob. 24.59PCh. 24 - Prob. 24.60PCh. 24 - Prob. 24.61PCh. 24 - Prob. 24.62PCh. 24 - Prob. 24.63PCh. 24 - Prob. 24.64PCh. 24 - Prob. 24.65PCh. 24 - Prob. 24.66PCh. 24 - Prob. 24.67PCh. 24 - Prob. 24.68PCh. 24 - Prob. 24.69PCh. 24 - Prob. 24.70PCh. 24 - Prob. 24.71PCh. 24 - Prob. 24.72PCh. 24 - Prob. 24.73PCh. 24 - Prob. 24.74PCh. 24 - Prob. 24.75PCh. 24 - Prob. 24.76PCh. 24 - Prob. 24.77PCh. 24 - Prob. 24.78PCh. 24 - Prob. 24.79PCh. 24 - Prob. 24.80PCh. 24 - Prob. 24.81PCh. 24 - Prob. 24.82PCh. 24 - Prob. 24.83PCh. 24 - Prob. 24.84PCh. 24 - Prob. 24.85PCh. 24 - Prob. 24.86PCh. 24 - Prob. 24.87PCh. 24 - Prob. 24.88PCh. 24 - Prob. 24.89PCh. 24 - Prob. 24.90PCh. 24 - Prob. 24.91PCh. 24 - Prob. 24.92PCh. 24 - Prob. 24.93PCh. 24 - Prob. 24.94PCh. 24 - Prob. 24.95PCh. 24 - Prob. 24.96PCh. 24 - Prob. 24.97PCh. 24 - Prob. 24.98PCh. 24 - Prob. 24.99PCh. 24 - Prob. 24.100PCh. 24 - Prob. 24.101PCh. 24 - Prob. 24.102PCh. 24 - Prob. 24.103PCh. 24 - Prob. 24.104PCh. 24 - Prob. 24.105PCh. 24 - Prob. 24.106PCh. 24 - Prob. 24.107PCh. 24 - Prob. 24.108PCh. 24 - Prob. 24.109PCh. 24 - Prob. 24.110PCh. 24 - Prob. 24.111PCh. 24 - Prob. 24.112PCh. 24 - Prob. 24.113PCh. 24 - Prob. 24.114PCh. 24 - Prob. 24.115PCh. 24 - Prob. 24.116PCh. 24 - Prob. 24.117PCh. 24 - Prob. 24.118PCh. 24 - Prob. 24.119PCh. 24 - Prob. 24.120PCh. 24 - Prob. 24.121PCh. 24 - Prob. 24.122PCh. 24 - Prob. 24.123PCh. 24 - Prob. 24.124PCh. 24 - Prob. 24.125PCh. 24 - Prob. 24.126PCh. 24 - Prob. 24.127PCh. 24 - Prob. 24.128PCh. 24 - Prob. 24.129PCh. 24 - Prob. 24.130PCh. 24 - Prob. 24.131PCh. 24 - Prob. 24.132PCh. 24 - Prob. 24.133PCh. 24 - Prob. 24.134PCh. 24 - Prob. 24.135PCh. 24 - Prob. 24.136PCh. 24 - Prob. 24.137PCh. 24 - Prob. 24.138PCh. 24 - Prob. 24.139PCh. 24 - Prob. 24.140PCh. 24 - Prob. 24.141PCh. 24 - Prob. 24.142PCh. 24 - Prob. 24.143PCh. 24 - Prob. 24.144P
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