chapter 7 answers

chapter 7 answers - See Sedan 7.]. a Solvingkv=cfork gives...

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Unformatted text preview: See Sedan 7.]. a Solvingkv=cfork gives 17:5 andremgnizlngthatll-LzEls'Ifields V B _ .1 12 A=3.00x10mms x10 pm=Lost 1.44x10 H2 m .30/ SezSect‘t'an 7.1 and Example 7.2. r (a) Salmng ).v = c for x gtves k = 5 . and recognizing that 1 Hz = 15" yields V l -| A; fling? :105 x to" mamas am 9‘8} x 10‘ s' (h) g = hv : (6.63 x 10'“ J - s) (9.33 x 10“ s") = 6.52 110'" J J 6.022x102‘3 photons lld x—-————-——x———= 393 kJ/mol (c) .‘E for on: mol of photons = 6.52 x 10'” photon 1 met 101 I 7.361 See Section 7.1 andEmnIple 7.1. 3 I -l 9 =3.002(10 ms 3"10 @gs‘slxmujt Solvin )w= forv 'ves = g C E! v 544mm 1m :- I V" (1,33) @5er 'u. if 77 _ 7 _ r 777 -I9 6 electmnsx 1.602»: 10 con] x10 won] =‘.00 pmul =4.00 “amps s l electron 1 can] 5 An electron is dislodged from a metal by collision with a single photon. having an energy of hvi provided that energy exceeds the threshold energy Hence, 2.50 x lo‘3 photons must be absorbed per second to produce 150 x 10‘3 electrons per second. ? mlcroamps = 2.50 x10” < 7J0 I See Senior: 7.2 andExanlpie Z3; £211., 4;; :Loenlc’ m"($=—L);2_zo4xtoém* x _ 1 22 52 "R "B 1 = 4.340 x to" m Solvmg for it yields J. = 6 7 _1 2. 304 x 10 m I This wavelength is equivalent to 434 run The radiation is in the visible portion of the electromagnetic spectmm. ( 7.46. Seaman ZJaadEur-Ipie' 7.4. h Soln'ng p=mv=— high/uh:— L mv -34 _ 2_-t (a)k:6.63x10 kgm s =9_s“o.{§,m (68 kg) [0 m-s") _ 6.63 x10“ kg-m: 45“ (b) - = 2.97 1 10*“ at (0.0500 kg)i44.7 m-s" 6.63 x 10'” ltgom2 ~s" (c) ,1: , =6.06 :10" m {9.11 x 10‘" kgkll x 10‘5 m-s'l “3+ on demon 1.4. (3) Since t mustbeatleastonelessthann.avalucof! =lisnotpossiblewlienn= 1‘ (b) n=4and£ =Zaxeassoclatedwlththe4dsubshcll. (c) n =Zandl=0massociatedwiththe255ubshell 4) Since allowed values for m! are all whole timbers from -£ to 4»: , a gal“: 9f ",1 = 1 is no! possible Wm [= 0. (c) n = 3 and E = 2 are associated with the 3d subshell. (D nzéand l =Oaxeassociatedwiththefissubshell. y ' 7.58 W See Section 7:4 andfigures 7.12. 7.14. 7.15. Ky 7, (a)n=33nd£=0massoctatedwith 38 O the 3s orbital. which is spherical. (b) The 3a,, is 45“ OH the x and y axes in the plane farmed by the utter- 3dxy section of the x and y axes. (c) The 411,. is "dumbbell" shaped along the y axis. I 168 ; Seeisen 11mm [£727. The wavelength of light absorbed when an electron moves from the n = 3 to n = 4 Bohr orbits of He‘ is equal to the wavelength 6f light that would be emitted as an eleetmn moves from the it = 4 te n i 3 Bohr orbits et’ He’. Substimungin:L=Zth[—l1—-iz] dnotlngthatZ=2forHe+gives A: n; itB 2 are 14232 1.097 x107 m") L--‘- =(2)2 1.097x10’ m‘1 4 37 :Zi133x105 m" ,t ‘ 31 x41 501mg forlyields h = = 4.683 x to" m 2.133 x 10“ m" This wavelength is equivalent to 468.8 tun. 7i72\ See Sening 7. 6. (a) He 15; Be 1512:2 Ne Isak2 Zp’i (tnN 152512131 1 _ l q 5 (c) H lsI Ll 15:25I El 15"25521:l I‘- Is 25‘2p \ 41.41 l at: dilly" Ad. (3,) Since I mustbcatlcastonclcsslhannjvflue ofl= l lsmtpfissiblewhenn= l. (b)n=4andl=2massociamdw1lhthc4dmbshell (c) n =23udt=03maswciatedwllhmelsmbshcll d) since allowed values for ml are all whole numbers finm —( to H , avaluc of m, = l is not possible when t = 0. (e) n = 3 and t = 2 are associated with the 3d subshell. (o n = 6 and t = 0 an: associated with the 6s subshell. (a)n=3and£ illmassociatedwim 35 C) the 3s orbital. which is sphencal. (1:) Th: 3d“. is 45° 06' the x and y axes in the' plane formed by :11: mm- 3dxy semen of the x and y axes. (c) The 4p, is “dumbbell” shaped along the y axis. 4p)’ m See 'eaion 7. J an Emmy: 7. 7. The wavelength of light absorbed when an electmn moves from th: :1 = 3 to n = 4 Bohr orbits och' is aqml to th: wavelength of light that would be sauna! as an electron moves fmm (h: n = 4 to n - 3 Bohr orbits och‘. Subsutuling in = Zth and noting marl a l {at He+ gives A n; as 42 _ 32 32 x41 = 2.133 x106 m" ,1 A =l2)2(1.097 x107 m1)[———]=(2)’(1.097x 107 m")[ Solnng for ,2. yiclds SL = = 4.688 x 10" m 2433 x10“ m“ This wavelength is equivalent to 4683 Inn. 7.72 ‘. See Section 7.6. (a) H: 15: B: 1.522;;2 N: 132251sz (b) N 1512522.:J 1 ~ (c) H 15‘ Li Islzs1 B Is‘Zlep' F 1s2 25‘ 2p’ \7 111; m lowest potential energy states correspond to n = l. 2. and 3 because the elements ctosm to me want". nucleat charge. -II .ll ; = 4.13 x 10'“ J E1: H’s—ZJL—J = 4.45 x10", J .2. 3 0'” E3 = 1 I = .242 :10", .t Nous: that the potential energy of an election in an atom is always negative. The reference point of infinite separation between the nucleus and the electron corresponds to zero potential energy. As the distance between the nucleus and the electron increases, the potential energy becomes less negative corresponding to higher energy El 512-165. 44 :- .1 - r _ 7 _6.63x10 kgm 5 [ml 1km‘36005 “imam meelmmifhl‘fasthall.k-___,___.___—X x 1 =. (t0220kg)(1mg) 1.6091011 10 m in: hr '3‘ 2 , -l . M 1m xflxfl= Ahma‘m Formemmifhrcurveball.l= I x 1.609kn1103m In: (.0220 “mam:— The uncertainty in the position ofthc too mi/hr fastball is (meta x 10-” m) = 3.37 x 10'" m and the uncertainty in the position of the 80 mith unveball is 096.42 x 10'“ tn) - 4.21 x 10‘” m. Hence, the uncertainty in the position of the fastball is ealctflated to be less than that for the curt-eta“. However. the uncertainty in the position ofbctth the fastball and the ctztveball is insignifitant. So this cannot be the reason why batten frequently ruins it muveball. ...
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This note was uploaded on 04/07/2008 for the course CHEM 125 taught by Professor Ellis during the Fall '07 term at NJIT.

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chapter 7 answers - See Sedan 7.]. a Solvingkv=cfork gives...

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