CHEMICAL PRINCIPLES (LL) W/ACCESS
CHEMICAL PRINCIPLES (LL) W/ACCESS
7th Edition
ISBN: 9781319421175
Author: ATKINS
Publisher: MAC HIGHER
bartleby

Concept explainers

Question
Book Icon
Chapter 1, Problem 1.31E

(a)

Interpretation Introduction

Interpretation:

Laser that can be used has to be determined among a high-intensity red ruby laser and a low-intensity violet GaN laser.

Concept Introduction:

Wavelength and frequency are inversely proportional to each other and the relationship between wavelength and frequency can be given as,

  ν=cλ

Here, ν is the frequency, λ is the wavelength and c is the speed of light

Planck’s equation,

    E = hν

Here, ν is the frequency

(a)

Expert Solution
Check Mark

Explanation of Solution

  • Calculate the energy of light from a high-intensity red ruby laser:

Wavelength of light from a high-intensity red ruby laser is 694 nm = 694×109m.

The frequency of light from a high-intensity red ruby laser is calculated by using the equation,

    ν=cλ=2.998×108m/s694×109m=4.32×1014s1

The frequency of light is 4.32×1014s1

The energy per photon of light is calculated,

  E=

Substituting the values to the above equation,

  E==6.626×1034J.s×4.32×1014s1=2.86×1019J

The energy per photon is 2.86×1019J.

  • Calculate the energy of light from a low-intensity violet GaN laser:

Wavelength of light from a low-intensity violet GaN laser is 405 nm = 405×109m.

The frequency of light from a low-intensity violet GaN laser is calculated by using the equation,

    ν=cλ=2.998×108m/s405×109m=7.40×1014s1

The frequency of light is 7.40×1014s1

The energy per photon of light is calculated,

  E=

Substituting the values to the above equation,

  E==6.626×1034J.s×7.40×1014s1=4.90×1019J

The energy per photon is 4.90×1019J.

Comparing the values it is clear that the energy of light from a low-intensity violet GaN laser is higher than that of a high-intensity red ruby laser. Hence, a low-intensity violet GaN laser is used since it will give enough energy for ejecting an electron.

(b)

Interpretation Introduction

Interpretation:

Kinetic energy of the electrons emitted has to be calculated.

Concept Introduction:

Photoelectric effect: It is a process where electrons get ejected when a ray of light having adequately high frequency hits on the surface of metal.

Kinetic energy of one ejected electron can be calculated using the given formula,

  12meν2 = hνΦ

Here, ν is the frequency of incident radiation.

12meν2 is the kinetic energy of one ejected electron

hν is the energy of the photons in the incident radiation.

Φ is the energy needed for ejecting an electron (work function).

(b)

Expert Solution
Check Mark

Answer to Problem 1.31E

Kinetic energy of the electrons emitted is 2.10×1020 J.

Explanation of Solution

Given information is shown below,

  Φ = 2.93 eV= 2.93 eV×1.602×1019 J1 eV = 4.69×1019 J

Here, a low-intensity violet GaN laser is used

The frequency of light from a low-intensity violet GaN laser is calculated by using the equation,

    ν=cλ=2.998×108m/s405×109m=7.40×1014s1

The energy per photon of light is calculated,

  E==6.626×1034J.s×7.40×1014s1=4.90×1019J

The kinetic energy of the electrons emitted can be calculated as follows,

  12meν2 = hνΦ= 4.90×1019J4.69×1019 J= 2.10×1020 J

Therefore, kinetic energy of the electrons emitted is 2.10×1020 J.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
eks.com/aleksogi/x/sl.exe/1o_u-IgNslkr7j8P3jH-IQs_pBanHhvTCeeBZbufuBYTI0Hz7m7D3ZS17Hd6m-HIl6n52njJN-TXdQA2X9yID-1SWQJTgnjARg30 111 States of Matter Understanding conceptual components of the enthalpy of solution 0/5 Ge A small amount of acetonitrile (CH, CN) is dissolved in a large amount of water. Imagine separating this process into the four stages sketched below. (These sketches show only a portion of the substances, so you can see the density and distribution of atoms and molecules in them.) CH,CN H₂O B 88 C Use these sketches to answer the questions in the table below. The enthalpy of solution AH is negative soln when CH3CN dissolves in water. Use this information to list the stages in order of increasing enthalpy. Would heat be absorbed or released if the system moved from Stage C to D? What force would oppose or favor the system moving from Stage C to D? Check all that apply. 1 absorbed O released neither absorbed nor released. none O ionic bonding force covalent bonding force…
In a system with an anodic overpotential, the variation of ŋ as a function of the current density: 1. at low fields is linear 2. at higher fields, it follows Tafel's law Find the range of current densities for which the overpotential has the same value as when calculated for cases 1 and 2 (maximum relative difference of 5% with respect to the behavior for higher fields). To which overpotential range does this correspond? Data: 10 = 1.5 mA cm², T = 300°C, ẞ = 0.64, R = 8.314 J K 1 mol‍¹ and F = 96485 C mol-1.
Indicate 10.6 with only one significant figure.

Chapter 1 Solutions

CHEMICAL PRINCIPLES (LL) W/ACCESS

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
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry In Focus
Chemistry
ISBN:9781337399692
Author:Tro, Nivaldo J.
Publisher:Cengage Learning,
Text book image
Physical Chemistry
Chemistry
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Wadsworth Cengage Learning,
Text book image
Fundamentals Of Analytical Chemistry
Chemistry
ISBN:9781285640686
Author:Skoog
Publisher:Cengage
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning