# chap42 - 42.1 a K J 10 38 1 3 eV J 10 60 1 eV 10 9 7 2 3 2...

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Unformatted text preview: 42.1: a) K) J 10 38 . 1 ( 3 ) eV J 10 60 . 1 )( eV 10 9 . 7 ( 2 3 2 2 3 23 19 4--- × × × = = ⇒ = k K T kT K K 1 . 6 = ⇒ T b) K. 600 , 34 ) K J 10 38 . 1 ( 3 ) eV J 10 60 . 1 ( ) eV 48 . 4 ( 2 23 19 = × × =-- T c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of 2 He (calculated in part (a)), so the typical collision at room temperature will be more than enough to break up . He 2 However, the thermal energy at 300 K is much less than the bond energy of 2 H , so we would expect it to remain intact at room temperature. 42.2: a) . eV . 5 4 1 2- =- = r e πε U b) . eV 2 . 4 ) eV 5 . 3 eV 3 . 4 ( eV . 5- =- +- 42.3: Let 1 refer to C and 2 to O. nm 1128 . , kg 10 656 . 2 , kg 10 993 . 1 26 2 26 1 = × = × =-- r m m ) carbon ( nm 0644 . 2 1 2 1 = + = r m m m r ) carbon ( nm 0484 . 2 1 1 2 = + = r m m m r b) ; m kg 10 45 . 1 2 46 2 2 2 2 1 1 ⋅ × = + =- r m r m I yes, this agrees with Example 42.2. 42.4: The energy of the emitted photon is , eV 10 01 . 1 5- × and so its frequency and wavelength are GHz 44 . 2 ) s J 10 63 . 6 ( ) eV J 10 60 . 1 )( eV 10 01 . 1 ( 34 19 5 = ⋅ × × × = =--- h E f . m 123 . Hz) 10 44 . 2 ( ) s m 10 00 . 3 ( λ 9 8 = × × = = f c This frequency corresponds to that given for a microwave oven. 42.5: a) From Example 42.2, 2 46 23 1 m kg 10 449 . 1 and J 10 674 . 7 meV 479 . ⋅ × = × = =-- I E s rad 10 03 . 1 2 gives and 12 1 2 2 1 × = = = = I E ω E K Iω K b) (carbon) s m 3 . 66 ) s rad 10 03 . 1 )( m 10 0644 . ( 12 9 1 1 1 = × × = =- ω r v (oxygen) s m 8 . 49 ) s rad 10 03 . 1 )( m 10 0484 . ( 12 9 2 2 2 = × × = =- ω r v c) s 10 6.10 2 12- × = = ω π T 42.6: a) J 10 083 . 2 ) eV 2690 . ( 2 1 2 1 20- × = = = ω E K s m 10 91 . 1 kg 10 139 . 1 ) J 10 083 . 2 ( 2 2 gives 2 1 3 26 2 r max 2 max r × = × × = = =-- m E v v m E b) According the Eq. 42.7 the spacing between adjacent vibrational energy levels is twice the ground state energy: . , ) 2 1 ( 1 hf ω E E E ω n E n n n = =- = ∆ + = + Thus, using the E ∆ specified in Example 42.3, it follows that its vibrational period is s. 10 54 . 1 ) eV J 10 60 . 1 )( eV 2690 . ( ) s J 10 63 . 6 ( 1 14 19 34--- × = × ⋅ × = ∆ = = E h f T c) The vibrational period is shorter than the rotational period. 42.7: a) 2 H Li H Li 2 r r m m m m r m I + = = eV. 10 53 . 7 J 10 20 . 1 m kg 10 69 . 3 ) s J 10 054 . 1 ( 4 4 )) 1 3 ( ) 3 ( ) 1 4 ( 4 ( 2 m kg 10 69 . 3 ) kg 10 67 . 1 kg 10 17 . 1 ( ) m 10 59 . 1 )( kg 10 67 . 1 )( kg 10 17 . 1 ( 3 21 2 47 2 34 2 2 3 4 2 47 27 26 2 10 27 26---------- × = × = ∆ ⇒ ⋅ × ⋅ × = ∆ ⇒ = +- + =- = ∆ ⋅ × = × + × × × × = E E I I E E E K K b) . m 10 66 . 1 λ J 10 20 . 1 s) m 10 00 . 3 )( s J 10 63 . 6 ( λ 4 21 8 34--- × = ⇒ × × ⋅ × = ∆ = E hc 42.8: Each atom has a mass m and is at a distance 2 L from the center, so the moment of inertia is . m kg...
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## This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap42 - 42.1 a K J 10 38 1 3 eV J 10 60 1 eV 10 9 7 2 3 2...

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