Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 7, Problem 7.74QP

(a)

Interpretation Introduction

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

Concept introduction: The displacement of an object with respect to time is known as velocity. Velocity shows inverse relation with the wavelength of an object. It is measured in meter per second.

To determine: The velocity of a given object.

(a)

Expert Solution
Check Mark

Answer to Problem 7.74QP

Solution

The velocity of an electron is 9.7×102m/s_ .

Explanation of Solution

Explanation

Given

Mass of an 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 wavelength of an electron is 750nm .

The conversion of nm to m is done as,

1nm=109m

Therefore, the conversion of 750nm to m is done as,

750nm=750×109m=7.5×107m

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.

Rearranging the above equation in terms of velocity (u) ,

u=hmλ

Also, 1J=1kg.m2/s2

Substitute the values of h , λ and m in the above equation to calculate the velocity.

u=6.626×1034kg.m2/s2.s(9.10938×1031kg)(7.5×107m)=6.626×1034kg.m2/s6.832×1037kg.m=969.89.7×102m/s_

Hence, the velocity of an electron is 9.7×102m/s_ .

(b)

Interpretation Introduction

To determine: The velocity of a given object.

(b)

Expert Solution
Check Mark

Answer to Problem 7.74QP

Solution

The velocity of a proton is 5.283×101m/s_ .

Explanation of Solution

Explanation

Given

Mass of a proton is 1.67262×1024g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 1.67262×1024g to kg is done as,

1.67262×1024g=1.67262×1024×103kg=1.67262×1027kg

The wavelength of a proton is 750nm .

The conversion of nm to m is done as,

1nm=109m

Therefore, the conversion of 750nm to m is done as,

750nm=750×109m=7.5×107m

The velocity of a proton is calculated by using the equation,

u=hmλ

Also, 1J=1kg.m2/s2

Substitute the values of h , λ and m in the above equation to calculate the velocity.

u=6.626×1034kg.m2/s2.s(1.67262×1027kg)(7.5×107m)=6.626×1034kg.m2/s1.254×1033kg.m=5.283×101m/s_

Hence, the velocity of a proton is 5.283×101m/s_ .

(c)

Interpretation Introduction

To determine: The velocity of a given object.

(c)

Expert Solution
Check Mark

Answer to Problem 7.74QP

Solution

The velocity of a neutron is 5.275×101m/s_ .

Explanation of Solution

Explanation

Given

Mass of a neutron is 1.67493×1024g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 1.67493×1024g to kg is done as,

1.67493×1024g=1.67493×1024×103kg=1.67493×1027kg

The wavelength of a neutron is 750nm .

The conversion of nm to m is done as,

1nm=109m

Therefore, the conversion of 750nm to m is done as,

750nm=750×109m=7.5×107m

The velocity of a neutron is calculated by using the equation,

u=hmλ

Also, 1J=1kg.m2/s2

Substitute the values of h , λ and m in the above equation to calculate the velocity.

u=6.626×1034kg.m2/s2.s(1.67493×1027kg)(7.5×107m)=6.626×1034kg.m2/s1.256×1033kg.m=5.275×101m/s_

Hence, the velocity of a neutron is 5.275×101m/s_ .

(d)

Interpretation Introduction

To determine: The velocity of a given object.

(d)

Expert Solution
Check Mark

Answer to Problem 7.74QP

Solution

The velocity of an alpha particle is 1.33×101m/s_ .

Explanation of Solution

Explanation

Given

Mass of an alpha particle is 6.64×1024g .

The conversion of g to kg is done as,

1g=103kg

Therefore the conversion of 6.64×1024g to kg is done as,

6.64×1024g=6.64×1024×103kg=6.64×1027kg

The wavelength of an alpha particle is 750nm .

The conversion of nm to m is done as,

1nm=109m

Therefore, the conversion of 750nm to m is done as,

750nm=750×109m=7.5×107m

The velocity of an alpha particle is calculated by using the equation,

u=hmλ

Also, 1J=1kg.m2/s2

Substitute the values of h , λ and m in the above equation to calculate the velocity.

u=6.626×1034kg.m2/s2.s(6.64×1027kg)(7.5×107m)=6.626×1034kg.m2/s4.98×1033kg.m=1.33×101m/s_

Hence, the velocity of an alpha particle is 1.33×101m/s_ .

Conclusion

  1. a. The velocity of an electron is 9.7×102m/s_ .
  2. b. The velocity of a proton is 5.283×101m/s_ .
  3. c. The velocity of a neutron is 5.275×101m/s_ .
  4. d. The velocity of an alpha particle is 1.33×101m/s_

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Chapter 7 Solutions

Chemistry

Ch. 7.8 - Prob. 11PECh. 7.9 - Prob. 12PECh. 7.9 - Prob. 13PECh. 7.10 - Prob. 14PECh. 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. 7.11VPCh. 7 - Prob. 7.12VPCh. 7 - Prob. 7.13VPCh. 7 - Prob. 7.14VPCh. 7 - Prob. 7.15VPCh. 7 - Prob. 7.16VPCh. 7 - Prob. 7.17VPCh. 7 - Prob. 7.18VPCh. 7 - Prob. 7.19VPCh. 7 - Prob. 7.20VPCh. 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. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 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. 7.127QPCh. 7 - Prob. 7.128QPCh. 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. 7.142APCh. 7 - Prob. 7.143APCh. 7 - Prob. 7.144APCh. 7 - Prob. 7.145APCh. 7 - Prob. 7.146APCh. 7 - Prob. 7.147APCh. 7 - Prob. 7.148AP
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