1131_PartUniversity Physics Solution

# 1131_PartUniversity Physics Solution - 42-12 Chapter 42(ii...

This preview shows pages 1–3. Sign up to view the full content.

42-12 Chapter 42 (ii) transition 1 0, 2 1 n n l l = = = = 2 2 2 (6 2) 2 E hf hf I I Δ = + = + = = 8 11 13 2.998 10 m/s 4.28 m 2( / 2 ) 2(3.960 10 Hz) 6.93 10 Hz c I f λ μ π × = = = + × + × = (iii) transition 2 1, 2 3 n n l l = = = = 2 2 3 (6 12) 2 E hf hf I I Δ = + = − + = = 8 11 13 2.998 10 m/s 4.40 m 3( / 2 ) 3(3.960 10 Hz) 6.93 10 Hz c I f λ μ π × = = = + × + × = E VALUATE : The vibrational energy change for the 1 0 n n = = transition is the same as for the 2 1 n n = = transition. The rotational energies are much smaller than the vibrational energies, so the wavelengths for all three transitions don’t differ much. 42.48. The sum of the probabilities is / F F 1 1 1 ( ) ( ) 1. 1 1 1 1 E kT E kT E kT E kT E kT e f E E f E E e e e e −Δ −Δ Δ −Δ −Δ + Δ + − Δ = + = + = + + + + 42.49. Since potassium is a metal we approximate F F0 . E E = 2 3 4 3 2 2 3 F 3 2 π n E m = = . 3 28 3 26 2 3 4 3 34 2 28 3 2 3 19 F 31 851kg m But the electron concentration 1.31 10 electron m 6.49 10 kg 3 (1.054 10 J s) (1.31 10 /m ) 3.24 10 J 2.03 eV. 2(9.11 10 kg) ρ n n m π E = = = × × × × = = × = × 42.50. I DENTIFY : The only difference between the two isotopes is their mass, which will affect their reduced mass and hence their moment of inertia. S ET U P : The rotational energy states are given by 2 ( 1) 2 E l l I = + = and the reduced mass is given by m r = m 1 m 2 /( m 1 + m 2 ). E XECUTE : (a) If we call m the mass of the H-atom, the mass of the deuterium atom is 2 m and the reduced masses of the molecules are H 2 (hydrogen): m r (H) = mm/ ( m + m ) = m /2 D 2 (deuterium): m r (D) = (2 m )(2 m )/(2 m + 2 m ) = m Using I = m r r 0 2 , the moments of inertia are I H = mr 0 2 /2 and I D = mr 0 2 . The ratio of the rotational energies is then ( ) ( ) 2 2 H H D 0 2 2 D H D 0 ( 1) /2 2 ( 1) / 2 2 l l I E I mr m E I l l I r + = = = = + = = . (b) The ratio of the vibrational energies is r H r D r r 1 2 (H) (D) 2 (H) /2 1 2 (D) k n m E m m E m m k n m + = = = = + = = . E VALUATE : The electrical force is the same for both molecules since both H and D have the same charge, so it is reasonable that the force constant would be the same for both of them. 41.51. I DENTIFY and S ET U P : Use the description of the bcc lattice in Fig.42.12c in the textbook to calculate the number of atoms per unit cell and then the number of atoms per unit volume. E XECUTE : (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells. So there are 1 8/8 2 + = atoms per unit cell. 28 3 9 3 2 4.66 10 atoms/m (0.35 10 m) n V = = × × (b) 2 / 3 2 / 3 4 / 3 2 F0 3 2 N E m V π = = In this equation N / V is the number of free electrons per 3 m . But the problem says to assume one free electron per atom, so this is the same as n / V calculated in part (a). 31 9.109 10 kg m = × (the electron mass), so 19 F0 7.563 10 J 4.7 eV E = × = E VALUATE : Our result for metallic lithium is similar to that calculated for copper in Example 42.8.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Molecules and Condensed Matter 42-13 42.52. (a) 2 tot 2 9 0 1 1 8 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern