Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 7, Problem 7.65QP

(a)

Interpretation Introduction

Interpretation: The wavelength of the given objects is to be calculated.

Concept introduction: Wavelength is defined as the equal distance between the two successive troughs and wave crests. It is denoted by a symbol lambda (λ) and is calculated in the unit of meter or nanometer. The wavelength of photons emitted is inversely proportional to the frequency of photon.

To determine: The wavelength of the given object.

(a)

Expert Solution
Check Mark

Answer to Problem 7.65QP

Solution

The wavelength of a muon is 10.82nm_ .

Explanation of Solution

Explanation

Given

Mass of muon (a subatomic particle) is 1.884×1025g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 1.884×1025g to kg is done as,

1.884×1025g=1.884×1025×103kg=1.884×1028kg

The velocity of particle is 325m/s .

The wavelength of a moving particle is given by the equation which is represented as,

λ=hmu

Where,

  • h is the Planck’s constant (6.626×1034J.s) .
  • u is the velocity in m/s .
  • m is the mass of the particle in kg .
  • λ is the wavelength of a particle in nm.

Also, 1J=1kg.m2/s2

Substitute the values of h , u and m in the above equation to calculate the wavelength.

λ=6.626×1034J.s(1.884×1028kg)(325m/s)=6.626×1034kg.m2/s2.s6.123×1026kg.m/s=1.082×108m

The wavelength of an object is 1.082×108m .

The conversion of m to nm is done as,

1m=109nm

Therefore the conversion of 1.082×108m to nm is done as,

1.082×108m=1.082×108×109nm=10.82nm_

Hence, the wavelength of a muon is 10.82nm_ .

(b)

Interpretation Introduction

To determine: The wavelength of the given object.

(b)

Expert Solution
Check Mark

Answer to Problem 7.65QP

Solution

The wavelength of an electron is 0.1796nm_ .

Explanation of Solution

Explanation

Given

Mass of electron is 9.10938×1028g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 9.10938×1028g to kg is done as,

9.10938×1028g=9.10938×1028×103kg=9.10938×1031kg

The velocity of electron is 4.05×106m/s .

The wavelength of a moving electron is given by the equation which is represented as,

λ=hmu

Where,

  • h is the Planck’s constant (6.626×1034J.s) .
  • u is the velocity in m/s .
  • m is the mass of an electron in kg .
  • λ is the wavelength of an electron in nm.

Also, 1J=1kg.m2/s2

Substitute the values of h , u and m in the above equation to calculate the wavelength.

λ=6.626×1034J.s(9.10939×1031kg)(4.05×106m/s)=6.626×1034kg.m2/s2.s3.689×1024kg.m/s=1.796×1010m

The wavelength of an electron is 1.796×1010m .

The conversion of m to nm is done as,

1m=109nm

Therefore the conversion of 1.796×1010m to nm is done as,

1.796×1010m=1.796×1010×109nm=0.1796nm_

Hence, the wavelength of an electron is 0.1796nm_ .

(c)

Interpretation Introduction

To determine: The wavelength of the given object.

(c)

Expert Solution
Check Mark

Answer to Problem 7.65QP

Solution

The wavelength of an athlete is 1.235×1027nm_ .

Explanation of Solution

Explanation

Given

Mass of athlete is 80kg .

A mile distance is running by athlete in 4 minutes.

Distance is 1 mile.

Time is 4 minutes.

The conversion of mile to m is done as,

1mile=1609.34m

Therefore conversion of 1 mile to m is done as,

1mile=1609.34m

The conversion of min to sec is done as,

1min=60sec

Therefore, the conversion of 4 min to sec is done as,

4min=4×60sec=240sec

The speed of athlete is calculated by using the formula,

Speed=DistanceTime

Substitute the value of distance and time in the above formula to calculate the speed of an athlete.

Speed=1609.34m240s=6.705m/s

The speed of an athlete is 6.705m/s .

The wavelength is given by the equation which is represented as,

λ=hmu

Where,

  • h is the Planck’s constant (6.626×1034J.s) .
  • u is the velocity in m/s .
  • m is the mass in kg .
  • λ is the wavelength in nm.

Also, 1J=1kg.m2/s2

Substitute the values of h , u and m in the above equation to calculate the wavelength.

λ=6.626×1034J.s(80kg)(6.705m/s)=6.626×1034kg.m2/s2.s536.44kg.m/s=1.235×1036m

The wavelength of an athlete is 1.235×1036m .

The conversion of m to nm is done as,

1m=109nm

Therefore the conversion of 1.235×1036m to nm is done as,

1.235×1036m=1.235×1036×109nm=1.235×1027nm_

Hence, the wavelength of an athlete is 1.235×1027nm_ .

(d)

Interpretation Introduction

To determine: The wavelength of the given object.

(d)

Expert Solution
Check Mark

Answer to Problem 7.65QP

Solution

The wavelength of an object is 3.681×1054nm_ .

Explanation of Solution

Explanation

Given

The mass of earth is 6.0×1027g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 6.0×1027g to kg is done as,

6.0×1027g=6.0×1027×103kg=6.0×1024kg

The velocity is 3.0×104m/s .

The wavelength is given by the equation which is represented as,

λ=hmu

Where,

  • h is the Planck’s constant (6.626×1034J.s) .
  • u is the velocity in m/s .
  • m is the mass in kg .
  • λ is the wavelength in nm.

Also, 1J=1kg.m2/s2

Substitute the values of h , u and m in the above equation to calculate the wavelength.

λ=6.626×1034J.s(6.0×1024kg)(3.0×104m/s)=6.626×1034kg.m2/s2.s18×1028kg.m/s=3.681×1063m

The wavelength is 3.681×1063m

The conversion of m to nm is done as,

1m=109nm

Therefore the conversion of 3.681×1063m to nm is done as,

3.681×1063m=3.681×1063×109nm=3.681×1054nm_

Hence, the wavelength is 3.681×1054nm_ .

Conclusion

  1. a. The wavelength of a muon is 10.82nm_ .
  2. b. The wavelength of an electron is 0.1796nm_ .
  3. c. The wavelength of an athlete is 1.235×1027nm_ .
  4. d. The wavelength of an object is 3.681×1054nm_ .

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Chapter 7 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 7.9 - Prob. 11PECh. 7.10 - Prob. 12PECh. 7.10 - Prob. 13PECh. 7.11 - Prob. 14PECh. 7.12 - Prob. 15PECh. 7 - Prob. 7.1VPCh. 7 - Prob. 7.2VPCh. 7 - Prob. 7.3VPCh. 7 - Prob. 7.4VPCh. 7 - Prob. 7.5VPCh. 7 - Prob. 7.6VPCh. 7 - Prob. 7.7VPCh. 7 - Prob. 7.8VPCh. 7 - Prob. 7.9VPCh. 7 - Prob. 7.10VPCh. 7 - Prob. 7.11QPCh. 7 - Prob. 7.12QPCh. 7 - Prob. 7.13QPCh. 7 - Prob. 7.14QPCh. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - Prob. 7.42QPCh. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Prob. 7.100QPCh. 7 - Prob. 7.101QPCh. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Prob. 7.112QPCh. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - Prob. 7.119QPCh. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Prob. 7.123QPCh. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Prob. 7.126QPCh. 7 - Prob. 7.127APCh. 7 - Prob. 7.128APCh. 7 - Prob. 7.129APCh. 7 - Prob. 7.130APCh. 7 - Prob. 7.131APCh. 7 - Prob. 7.132APCh. 7 - Prob. 7.133APCh. 7 - Prob. 7.134APCh. 7 - Prob. 7.135APCh. 7 - Prob. 7.136APCh. 7 - Prob. 7.137APCh. 7 - Prob. 7.138APCh. 7 - Prob. 7.139APCh. 7 - Prob. 7.140APCh. 7 - Prob. 7.141APCh. 7 - Prob. 7.142AP
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