Foundations of Materials Science and Engineering
Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
bartleby

Concept explainers

Question
Book Icon
Chapter 2.9, Problem 48AAP

(a)

To determine

The wrong information from given set of information and reason for it.

(a)

Expert Solution
Check Mark

Answer to Problem 48AAP

The values of ,m and name are wrong in given set of information.

Explanation of Solution

Write the expression for subsidiary quantum number.

    =n1                                                                                (I)

Here, the subsidiary quantum number is and principal quantum number is n.

Write the expression for magnetic quantum number.

    m=21                                                                           (II)

Here, the magnetic quantum number is m.

Conclusion:

Substitute 3 for n in Equation (I).

    =31=2

Substitute 2 for in equation (II).

    m=(2×2)+1m=5

Since the value of =2 is n=3 and, therefore the name of the orbit is 3d.

Thus, from the given set of information the values of ,m and name are wrong. This is because for principal quantum number n=3, the value of subsidiary quantum number will be 2, the value of magnetic quantum number will be 5 and will be written as 2,1,0,+1,+2.and the name of the orbit will be 3d as =0 represents s sub shell, =1 represents p sub shell, =2 represents d sub shell and =3 represents f sub shell.

(b)

To determine

The wrong information from given set of information and reason for it.

(b)

Expert Solution
Check Mark

Answer to Problem 48AAP

The values of ,m and name are wrong in given set of information.

Explanation of Solution

Conclusion:

Substitute 2 for n in Equation (I).

    =21=1

Substitute 1 for in equation (II).

    m=(2×1)+1m=3

Since the value of =1 is n=2 and, therefore the name of the orbit is 2p.

Thus, from the given set of information the values of ,m and name are wrong. This is because for principal quantum number n=2, the value of subsidiary quantum number will be 1, the value of magnetic quantum number will be 3 and will be written as 1,0,+1.and the name of the orbit will be 2p as =0 represents s sub shell, =1 represents p sub shell, =2 represents d sub shell and =3 represents f sub shell.

(c)

To determine

The wrong information from given set of information and reason for it.

(c)

Expert Solution
Check Mark

Answer to Problem 48AAP

The values of ,m and name are wrong in given set of information.

Explanation of Solution

Conclusion:

Substitute 3 for n in Equation (I).

    =31=2

Substitute 2 for in Equation (II).

    m=(2×2)+1m=5

Since the value of =2 is n=3 and, therefore the name of the orbit is 3d.

Thus, from the given set of information the values of ,m and name are wrong. This is because for principal quantum number n=3, the value of subsidiary quantum number will be 2, the value of magnetic quantum number will be 5 and will be written as 2,1,0,+1,+2.and the name of the orbit will be 3d as =0 represents s sub shell, =1 represents p sub shell, =2 represents d sub shell and =3 represents f sub shell.

(d)

To determine

The wrong information from given set of information and reason for it.

(d)

Expert Solution
Check Mark

Answer to Problem 48AAP

The given set of information the values of ,m and name are wrong.

Explanation of Solution

Conclusion:

Substitute 3 for n in Equation (I).

    =31=2

Substitute 2 for in Equation (II).

    m=(2×2)+1m=5

Since the value of =2 is n=3 and, therefore the name of the orbit is 3d.

Thus, from the given set of information the values of ,m and name are wrong. This is because for principal quantum number n=3, the value of subsidiary quantum number will be 2, the value of magnetic quantum number will be 5 and will be written as 2,1,0,+1,+2.and the name of the orbit will be 3d as =0 represents s sub shell, =1 represents p sub shell, =2 represents d sub shell and =3 represents f sub shell.

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
8 mm- Top view -200 mm-180 mm- D B B 12 mm Side view B -8 mm D PROBLEM 1.56 In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter is to be used at A. Assuming that all other specifications remain unchanged, determine the allowable load P if an overall factor of safety of 3.0 is desired. PROBLEM 1.55 In the structure shown, an 8- mm-diameter pin is used at A, and 12-mm- diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. 20 mm P 8 mm- 12 mm- Front view
Where on the beam below is the Maximum Deflection likely to occur? 2P A "ती Point A Point B Point C Point D Point B or Point D ८ B प
Sign in ||! PDE 321 proje X IMB321 PDF Lecture 5 X PDF Planet Ec X PDF Planet Ec X PDF PEABWX PDF meeting x PDF GSS Quo X PDF File C:/Users/KHULEKANI/Downloads/CIVE%20281%20Ass-2.pdf Draw | | All | a | Ask Copilot + 1 of 7 | D SOLUTION B PROBLEM 12.16 Block 4 has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μ, = 0.20 H = 0.15. Knowing that P = 50 N→, determine (a) the acceleration of block B, (b) the tension in the cord. Constraint of cable: 2x + (x-x1) = x + x = constant. a+ag = 0, or aB = -a Assume that block A moves down and block B moves up. Block B: +/ΣF, = 0: NAB - WB cos 0 = 0 =ma: -T+μN + Wsin = We as g + ΣΕ We Eliminate NAB and aB- NAB B Nas HN UNA A NA -T+W(sin+μcоsе) = WB- g VD"M- g Block A: +/ΣF, = 0: NA-NAB - W₁cos + Psinė = 0 N₁ = N AB+W cose - Psin = (WB+WA)cose - Psinė ΣF=ma -T+Wsino-FAB-F + Pcos = CIVE 281 X + Ждал g Q | го || حالم ☑

Chapter 2 Solutions

Foundations of Materials Science and Engineering

Ch. 2.9 - Prob. 11KCPCh. 2.9 - Prob. 12KCPCh. 2.9 - Prob. 13KCPCh. 2.9 - Prob. 14KCPCh. 2.9 - Prob. 15KCPCh. 2.9 - Prob. 16KCPCh. 2.9 - Prob. 17KCPCh. 2.9 - Describe the terms (a) metallic radius. (b)...Ch. 2.9 - Prob. 19KCPCh. 2.9 - Prob. 20KCPCh. 2.9 - Prob. 21KCPCh. 2.9 - Prob. 22KCPCh. 2.9 - Prob. 23KCPCh. 2.9 - Prob. 24KCPCh. 2.9 - Describe the properties (electrical, mechanical,...Ch. 2.9 - Prob. 26KCPCh. 2.9 - Prob. 27KCPCh. 2.9 - Prob. 28KCPCh. 2.9 - The diameter of a soccer ball is approximately...Ch. 2.9 - Each quarter produced by the U.S. mint is made up...Ch. 2.9 - Sterling silver contains 92.5 wt% silver and 7.5...Ch. 2.9 - Prob. 32AAPCh. 2.9 - Prob. 33AAPCh. 2.9 - Prob. 34AAPCh. 2.9 - Prob. 35AAPCh. 2.9 - Prob. 36AAPCh. 2.9 - Prob. 37AAPCh. 2.9 - Prob. 38AAPCh. 2.9 - Prob. 39AAPCh. 2.9 - Prob. 40AAPCh. 2.9 - Prob. 41AAPCh. 2.9 - Prob. 42AAPCh. 2.9 - Prob. 43AAPCh. 2.9 - Prob. 44AAPCh. 2.9 - Prob. 45AAPCh. 2.9 - Prob. 46AAPCh. 2.9 - Prob. 47AAPCh. 2.9 - Prob. 48AAPCh. 2.9 - Prob. 49AAPCh. 2.9 - Prob. 50AAPCh. 2.9 - Write the electron configurations of the following...Ch. 2.9 - Prob. 52AAPCh. 2.9 - Prob. 53AAPCh. 2.9 - Prob. 54AAPCh. 2.9 - Prob. 55AAPCh. 2.9 - Prob. 56AAPCh. 2.9 - Prob. 57AAPCh. 2.9 - Prob. 58AAPCh. 2.9 - Prob. 59AAPCh. 2.9 - Prob. 60AAPCh. 2.9 - Prob. 61AAPCh. 2.9 - Prob. 62AAPCh. 2.9 - Prob. 63AAPCh. 2.9 - For each bond in the following series of bonds,...Ch. 2.9 - Prob. 65AAPCh. 2.9 - Prob. 66AAPCh. 2.9 - Prob. 67AAPCh. 2.9 - Prob. 68AAPCh. 2.9 - Prob. 69SEPCh. 2.9 - Most modern scanning electron microscopes (SEMs)...Ch. 2.9 - Prob. 71SEPCh. 2.9 - Of the noble gases Ne, Ar, Kr, and Xe, which...Ch. 2.9 - Prob. 73SEPCh. 2.9 - Prob. 74SEPCh. 2.9 - Prob. 75SEPCh. 2.9 - Prob. 76SEPCh. 2.9 - Prob. 77SEPCh. 2.9 - Prob. 78SEPCh. 2.9 - Prob. 79SEPCh. 2.9 - Prob. 80SEPCh. 2.9 - Silicon is extensively used in the manufacture of...Ch. 2.9 - Prob. 82SEPCh. 2.9 - Prob. 83SEPCh. 2.9 - Prob. 84SEPCh. 2.9 - Prob. 85SEPCh. 2.9 - Prob. 86SEPCh. 2.9 - Prob. 87SEPCh. 2.9 - Prob. 88SEPCh. 2.9 - Prob. 89SEPCh. 2.9 - Prob. 90SEPCh. 2.9 - Prob. 91SEPCh. 2.9 - Prob. 92SEP
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Electrical Transformers and Rotating Machines
Mechanical Engineering
ISBN:9781305494817
Author:Stephen L. Herman
Publisher:Cengage Learning