ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 1, Problem 1B.15E

(a)

Interpretation Introduction

Interpretation:

The wavelength of the ejected electron having wavelength of 3.6×103km.s1 has to be calculated.

Concept Introduction:

de Broglie relation:

Mathematically, de Broglie relation can be represented as given below.

  λ=hmv

Where, h is the Planck’s constant, λ is the wavelength of radiation, m is the mass of particle and v is the velocity of the particle.

(a)

Expert Solution
Check Mark

Answer to Problem 1B.15E

The wavelength of the ejected electron having wavelength of 3.6×103km.s1 is 2.02×10-10m.

Explanation of Solution

Given that, the velocity of electron is 3.6×103km.s1.  The mass of electron is 9.109×1031kg.

The de Broglie relation is given below.

  λ=hmv

By plugging all data in the above equation, the value of wavelength of the electron can be calculated.

  λ=hmv=6.626×1034J.s(9.109×1031kg)×(3.6×106m/s)λ=2.02×1010×kg.m2.s2.skg.m.s1λ=2.02×10-10m

Therefore, the wavelength of the ejected electron having wavelength of 3.6×103km.s1 is 2.02×10-10m.

(b)

Interpretation Introduction

Interpretation:

The energy required to remove the electron from the metal has to be calculated.

Concept Introduction:

Energy of a photon can be expressed mathematically as given below.

  E=hν

Where, h is the Planck’s constant, ν is the frequency of radiation.

(b)

Expert Solution
Check Mark

Answer to Problem 1B.15E

The energy required to remove the electron from the metal is 1.66×10-17J.

Explanation of Solution

Given that, the frequency of the radiation is 2.50×1016Hz.  The corresponding energy of a photon can be calculated as given below.

  E=hν=(6.626×1034J.s)×(2.50×1016s1)=1.66×10-17J.

Therefore, the energy required to remove the electron from the metal is 1.66×10-17J.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the radiation that caused photoejection of electron with a velocity of 3.6×103km.s1 has to be calculated.

Concept Introduction:

Photoelectric effect:

If the energy of the photon is greater than work function, then an electron can be ejected with a kinetic energy, Ek=12mev2.  An equation which represents the kinetic energy is given below.

  Ek=hνΦ

Where, Ek is the kinetic energy, h is the Planck’s constant, ν is the frequency of radiation and Φ is the work function.

(c)

Expert Solution
Check Mark

Answer to Problem 1B.15E

The wavelength of the radiation that caused photoejection of electron is 8.8nm.

Explanation of Solution

Given that, the speed of an electron that is emitted from the surface of a sample of chromium metal by a photon is 3.6×103km.s1.

The expression of kinetic energy is given below.

  Ek=12mev2.......(1)

Where, me is the mass of electron and v is the speed of electron.

By plugging all data in the above equation, the value of kinetic energy can be calculated.

  Ek=12mev2Ek=12×(9.109×1031kg)×(3.6×106m.s1)2Ek=5.9×10-18kg.m2.s-2=5.9×10-18J.

The energy of incoming photon can be calculated as given below.

  E(incomingphoton)=Ek+hν..........(2)[Φ=hν]

By plugging all data in the above equation, the energy of incoming photon can be calculated.

  E(incomingphoton)=Ek+hνE(incomingphoton)=Ek+hν=(5.9×10-18J)+(1.66×1017J)E(incomingphoton)=2.25×10-17Jhcλ=2.25×10-17Jλ=hc2.25×10-17J=(6.626×1034J.s)×(3×108m/s)2.25×10-17J=8.8×109m=8.8nm.

Therefore, the wavelength of the radiation that caused photoejection of electron is 8.8nm.

(d)

Interpretation Introduction

Interpretation:

The kind of electromagnetic radiation used has to be determined.

Concept Introduction:

The wavelengths of electromagnetic radiation and their corresponding frequencies are given below in the table.

Radiation typeFrequency / (1014Hz)Wavelength / (nm)Energy of photon / (1019J)
x-rays and γ- rays1033103
Ultraviolet8.63505.7
Visible light
Violet7.14204.7
Blue6.44704.2
Green5.75303.8
Yellow5.25803.4
Orange4.86203.2
Red4.37002.8
Infrared3.010002.0
Microwaves and radio waves1033×106103

(d)

Expert Solution
Check Mark

Explanation of Solution

The wavelength of the radiation that caused photoejection of electron is 8.8nm.

Therefore, the kind of electromagnetic radiation used is X- rays.

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Chapter 1 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 1 - Prob. 1A.8ECh. 1 - Prob. 1A.9ECh. 1 - Prob. 1A.10ECh. 1 - Prob. 1A.11ECh. 1 - Prob. 1A.12ECh. 1 - Prob. 1A.13ECh. 1 - Prob. 1A.14ECh. 1 - Prob. 1A.15ECh. 1 - Prob. 1A.16ECh. 1 - Prob. 1A.17ECh. 1 - Prob. 1B.1ASTCh. 1 - Prob. 1B.1BSTCh. 1 - Prob. 1B.2ASTCh. 1 - Prob. 1B.2BSTCh. 1 - Prob. 1B.3ASTCh. 1 - Prob. 1B.3BSTCh. 1 - Prob. 1B.4ASTCh. 1 - Prob. 1B.4BSTCh. 1 - Prob. 1B.5ASTCh. 1 - Prob. 1B.5BSTCh. 1 - Prob. 1B.1ECh. 1 - Prob. 1B.2ECh. 1 - Prob. 1B.3ECh. 1 - Prob. 1B.4ECh. 1 - Prob. 1B.5ECh. 1 - Prob. 1B.6ECh. 1 - Prob. 1B.7ECh. 1 - Prob. 1B.8ECh. 1 - Prob. 1B.9ECh. 1 - Prob. 1B.10ECh. 1 - Prob. 1B.11ECh. 1 - Prob. 1B.12ECh. 1 - Prob. 1B.13ECh. 1 - Prob. 1B.14ECh. 1 - Prob. 1B.15ECh. 1 - Prob. 1B.16ECh. 1 - Prob. 1B.17ECh. 1 - Prob. 1B.18ECh. 1 - Prob. 1B.19ECh. 1 - Prob. 1B.21ECh. 1 - Prob. 1B.22ECh. 1 - Prob. 1B.23ECh. 1 - Prob. 1B.24ECh. 1 - Prob. 1B.25ECh. 1 - Prob. 1B.26ECh. 1 - Prob. 1B.27ECh. 1 - Prob. 1B.28ECh. 1 - Prob. 1C.1ASTCh. 1 - Prob. 1C.1BSTCh. 1 - Prob. 1C.1ECh. 1 - Prob. 1C.2ECh. 1 - Prob. 1C.3ECh. 1 - Prob. 1C.7ECh. 1 - Prob. 1D.1ASTCh. 1 - Prob. 1D.1BSTCh. 1 - Prob. 1D.2ASTCh. 1 - Prob. 1D.2BSTCh. 1 - Prob. 1D.1ECh. 1 - Prob. 1D.2ECh. 1 - Prob. 1D.3ECh. 1 - Prob. 1D.4ECh. 1 - Prob. 1D.5ECh. 1 - Prob. 1D.6ECh. 1 - Prob. 1D.7ECh. 1 - Prob. 1D.9ECh. 1 - Prob. 1D.10ECh. 1 - Prob. 1D.11ECh. 1 - Prob. 1D.12ECh. 1 - Prob. 1D.13ECh. 1 - Prob. 1D.14ECh. 1 - Prob. 1D.15ECh. 1 - Prob. 1D.16ECh. 1 - Prob. 1D.17ECh. 1 - Prob. 1D.18ECh. 1 - Prob. 1D.19ECh. 1 - Prob. 1D.20ECh. 1 - Prob. 1D.21ECh. 1 - Prob. 1D.22ECh. 1 - Prob. 1D.23ECh. 1 - Prob. 1D.24ECh. 1 - Prob. 1D.25ECh. 1 - Prob. 1D.26ECh. 1 - Prob. 1E.1ASTCh. 1 - Prob. 1E.1BSTCh. 1 - Prob. 1E.2ASTCh. 1 - Prob. 1E.2BSTCh. 1 - Prob. 1E.1ECh. 1 - Prob. 1E.2ECh. 1 - Prob. 1E.3ECh. 1 - Prob. 1E.4ECh. 1 - Prob. 1E.5ECh. 1 - Prob. 1E.7ECh. 1 - Prob. 1E.8ECh. 1 - Prob. 1E.9ECh. 1 - Prob. 1E.10ECh. 1 - Prob. 1E.11ECh. 1 - Prob. 1E.12ECh. 1 - Prob. 1E.13ECh. 1 - Prob. 1E.14ECh. 1 - Prob. 1E.15ECh. 1 - Prob. 1E.16ECh. 1 - Prob. 1E.17ECh. 1 - Prob. 1E.18ECh. 1 - Prob. 1E.19ECh. 1 - Prob. 1E.20ECh. 1 - Prob. 1E.21ECh. 1 - Prob. 1E.22ECh. 1 - Prob. 1E.23ECh. 1 - Prob. 1E.24ECh. 1 - Prob. 1E.25ECh. 1 - Prob. 1E.26ECh. 1 - Prob. 1F.1ASTCh. 1 - Prob. 1F.1BSTCh. 1 - Prob. 1F.2ASTCh. 1 - Prob. 1F.2BSTCh. 1 - Prob. 1F.3BSTCh. 1 - Prob. 1F.1ECh. 1 - Prob. 1F.2ECh. 1 - Prob. 1F.3ECh. 1 - Prob. 1F.4ECh. 1 - Prob. 1F.5ECh. 1 - Prob. 1F.6ECh. 1 - Prob. 1F.7ECh. 1 - Prob. 1F.8ECh. 1 - Prob. 1F.10ECh. 1 - Prob. 1F.11ECh. 1 - Prob. 1F.12ECh. 1 - Prob. 1F.13ECh. 1 - Prob. 1F.14ECh. 1 - Prob. 1F.15ECh. 1 - Prob. 1F.17ECh. 1 - Prob. 1F.18ECh. 1 - Prob. 1F.19ECh. 1 - Prob. 1F.22ECh. 1 - Prob. 1.1ECh. 1 - Prob. 1.2ECh. 1 - Prob. 1.3ECh. 1 - Prob. 1.9ECh. 1 - Prob. 1.10ECh. 1 - Prob. 1.11ECh. 1 - Prob. 1.12ECh. 1 - Prob. 1.13ECh. 1 - Prob. 1.14ECh. 1 - Prob. 1.15ECh. 1 - Prob. 1.17ECh. 1 - Prob. 1.19ECh. 1 - Prob. 1.21ECh. 1 - Prob. 1.22ECh. 1 - Prob. 1.23ECh. 1 - Prob. 1.24ECh. 1 - Prob. 1.25ECh. 1 - Prob. 1.27ECh. 1 - Prob. 1.28ECh. 1 - Prob. 1.31E
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