Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 6, Problem 103AP
Interpretation Introduction

Interpretation:

The binding energy of each metal, and the metal with highest binding energy is to be predicted; and the metal thatejects electron when photon is fired on it is to be determined.

Concept introduction:

The energy of a photon can be expressed as follows:

E=hcλ 

Here, E is the energy of photon, h is Planck’s constant (6.63×1034 Js), c is the speed of light (3.0×108 m/s), and λ is the wavelength.

The relationship between wavelength and frequency is as follows:

c=λ×ν 

Here, c is the speed of light (3.0×108 m/s), λ is the wavelength, and ν  is frequency.

The relationship between binding energy, kinetic energy, and energy of ejected photon is as follows:

E=hν=W+K. E

Here, E is the energy of ejected photon, h is Planck’s constant, ν is frequency of radiation falling on the metal, W is the binding energy, and K. E. is the kinetic energy of the ejected photon.

Expert Solution & Answer
Check Mark

Answer to Problem 103AP

Solution:

(a)

The binding energy of metals A, B, and C is 3.4×1019 J, 5.6×1019 J, and 6.6×1019 J, respectively. Metal C has the highest binding energy.

(b)

An electron can be ejected when a radiation of minimum frequency falls on the metal surface. Therefore, on allowing a radiation of wavelength 333 nmto fall on metal Ametal B, they will eject electrons.

Explanation of Solution

(a)

Given information: The frequency of metal A (νA) is 5.2×1014 s1.

The frequency of metal B (νB) is 8.5×1014 s1.

The frequency of metal C (νC) is 9.9×1014 s1.

Consider the binding energy for metal A, B, and C tobe WA, WB, and WC, respectively.

In the photoelectric effect, the minimum frequency (or maximum wavelength) that is required to dislodge an electron from the metal surface is called threshold energy. The extra energy is stored in the electron as kinetic energy (K. E.). If kinetic energy becomes zero and the binding energy is equal to the energy of the ejected photon, that is,

W=hν

Now, the binding energy of electron for each metal can be evaluated as shown below:

For metal A:

WA=hνA=(6.63×1034 Js)(5.2×1014 s1)=3.4×1019 J

For metal B:

WB=hνB=(6.63×1034 Js)(8.5×1014 s1)=5.6×1019 J

For metal C:

WC=hνC=(6.63×1034 Js)(9.9×1014 s1)=6.6×1019 J

Therefore, the binding energy of metal C is highest.

(b)

Given information:

The wavelength of photon is 333 nm.

The frequency of metal A (νA) is 5.2×1014 s1.

The frequency of metal B (νB) is 8.5×1014 s1.

The frequency of metal C (νC) is 9.9×1014 s1.

The frequency associated with 333 nm wavelength can be evaluated as follows:

c=λ×ν3.0×108 m/s=333 nm×νν=3.0×108 m/s333 nm(1 nm1×109 m)=9.0×1014 s1

So, the frequency is 9.0×1014 s1.

Since the electron can be ejected when a radiation of minimum frequency falls on the metal surface;therefore, on allowing a radiation of wavelength 333 nm to fall on metal A and metal B, they will eject electrons.

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Chapter 6 Solutions

Chemistry

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