
The principle that summarises that no two electrons having the same space can have same quantum numbers.

Answer to Problem 1CQ
The Pauli Exclusion Principle states that no two electrons occupy the same space can have the same quantum numbers.
Explanation of Solution
The set of values which describes the electron state including the orientation of an electron, its distance from the nucleus and type of orbital, where the electron is found and its spin is known as quantum number.
There are four types of quantum number:
- Principal Quantum number
- Azimuthal Quantum number
- Magnetic Quantum number
- Spin Quantum number
The quantum number that is assigned to all electrons in an atom for the description of state of those electrons is known as Principal quantum number.
The quantum number which describes the shape of the orbital in which electron is present is known as Azimuthal quantum number.
The quantum number which describes the levels of energy in a subshell is known as Magnetic quantum number.
The quantum number which describes spin of electrons (up or down) is known as Spin quantum number.
Bonding energy depends upon the quantum
Pauli Exclusion Principle: Two electron of same atom cannot have same quantum numbers.
Let us take the example of
Conclusion:
The Pauli Exclusion Principle summarises that two electrons occupying the same space cannot possibly have the same quantum number as a whole.
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Chapter 2 Solutions
Materials Science And Engineering Properties
- A study reports data on the effects of the drug tamoxifen on change in the level of cortisol-binding globulin (CBG) of patients during treatment. With age = x and ACBG = y, summary values are n = 26, Σx, = 1612, Σ(x, - x)² = 3756.96, Σy, = 281.9, Σ(y, - y)² = 465.34, and Ex,y,= 16,745. (a) Compute a 90% CI for the true correlation coefficient p. (Round your answers to four decimal places.) (b) Test Hop=-0.5 versus H: p< -0.5 at level 0.05. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. ◇ Reject Ho. There is no evidence that p < -0.5. ○ Fail to reject Ho. There is evidence that p < -0.5. Reject Ho. There is evidence that p < -0.5. Fail to reject Ho. There is no evidence that p < -0.5. (c) In a regression analysis of y on x, what proportion of variation in change of cortisol-binding globulin level could be explained by variation in…arrow_forwardFor the frame and loading shown, determine the reactions at A and C. 24 Last 2 student ID+50 lbs 24 A 3 in. B A=Last 2 student ID+10 Inch B=Last 2 student ID+40 Inch A B Darrow_forwardThe figures below shows the framing plan and section of a reinforced concrete floor system. Floor beams are shown as dotted lines. The weight of the ceiling and floor finishing is 6 psf, that of the mechanical and electrical systems is 7 psf, and the weight of the partitions is 180 psf. The floor live load is 105 psf. The 7 in. thick slab exterior bay (S-1) is reinforced with #5 rebars @ 10 in. o.c. as the main positive reinforcement at the mid span, and #4 @ 109 in. for the shrinkage and temperature reinforcement. The panel is simply supported on the exterior edge and monolithic with the beam at the interior edge. Check the adequacy of the slab. Use the ACI moment coefficients. fc’ = 6,000 psi and fy = 60,000 psi. The slab is in an interior location. Hint: • Estimate total dead load. Find factored maximum positive bending moment in the end span. • Find design positive moment capacity. • Compare and determine adequacy, including safety and economy. C Darrow_forward
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- A simply supported rectangular prestressed concrete beam, of span 13 m and its cross section as shown in figure, is carrying a live load equals to 30 kN/m in addition to its weight, compute the following stresses and compare it with ACI allowable stress: a) Bottom fiber stress at support in initial stage. b) Top fiber stress at mid span in final stage. Use y = 24 kN/m³, As = 600 mm², initial stress of the prestressed steel = 1200 MPa, total losses is 20%, fci = 22 MPa, and fo' = 28 MPa 800 mm 3 As 400 mm As 340 mm 340 mmarrow_forwardA study of the properties of metal plate-connected trusses used for roof support yielded the following observations on axial stiffness index (kips/in.) for plate lengths 4, 6, 8, 10, and 12 in: 4: 320.2 409.5 311.0 326.5 316.8 349.8 309.7 6: 401.1 347.2 361.0 404.5 331.0 348.9 381.7 8: 395.4 366.2 351.0 357.1 409.9 367.3 382.0 10: 356.7 452.9 461.4 433.1 410.6 384.2 362.6 12: 415.4 441.8 419.9 410.7 473.4 441.2 465.8 USE SALT Does variation in plate length have any effect on true average axial stiffness? State the relevant hypotheses using analysis of variance. O Ho M1 M2 M3 = μ4=μ5 Ha all five μ's are unequal Ho: M1 M2 M3 #44 #μ5 Ha all five μ's are equal Ho: M₁ μ2 43 #44 #μ5 H₁: at least two μ,'s are equal O Ho M1 M2 M3 = μ4=μ5 Ha at least two μ's are unequal Test the relevant hypotheses using analysis of variance with a = 0.01. Display your results in an ANOVA table. (Round your answers to two decimal places.) Degrees of Sum of Mean Source Squares Squares f freedom Treatments Error…arrow_forward1-Define a cartesion system 2 identify the structure's supports and the type of structure (2D) or 3D 3-If the structure has more than one element dismember the structure and draw free body diagram(show all actions and reactions) on each element independently 4- Determine the type of suports 5- show the unkown supports reactions with any assumed direction but you cannot change the assumed force direction once you dicede 6-In a common joint, you can dicide on the force direction in one element, however, in the other one you need to follow the Newton'ns 3rd law and shoe the opposite direction 7- if you have multiple actions forces in the system, find force components for each foce independently use Sin/Cos/Tan functions to find forces components in two perpendicular directions 8- Add forces in each direction since they are paralled forces Rx=fx Ry=fy Rz= fyarrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage LearningConstruction Materials, Methods and Techniques (M...Civil EngineeringISBN:9781305086272Author:William P. Spence, Eva KultermannPublisher:Cengage Learning

