KeyHW1 - 1—7. Sirius, one of the hottest known stars, has...

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Unformatted text preview: 1—7. Sirius, one of the hottest known stars, has approximately a blackbody spectrum with Am“ = 260 nm. Estimate the surface temperature of Sirius. From Equation 1.4, 2.90 x 10-3 m-K 2.90 x 10‘3 m-K 4 T=—————————=1.12x10 K )t 260 X 104 m max 1—11. Calculate the number of photons in a 2.00 In] light pulse at (a) 1.06 am, (b) 537 nm, and (c) 266 nm. hc :hvz— )\. photon _ (6.626 X 10‘34 J-s)(2.998 X 108 ms”) _ 1.06 x 10‘6 m = 1.87 x 10*19 J -photon_1 Since 2.00 m] of energy are contained in the light pulse, 2.00 x 10*3 J .m . 1 = 1.UI X 1U puutuua N b 01 hotonS = ————‘—‘— um er p 1.87 x 10'”J-photon‘ Parts (b) and (c) are done in the same manner to find b. 5.41 X 1015 photons c. 2.68 X 1015 photons 1—16. The threshold wavelength for potassium metal is 564 nm. What is its work function? What is the kinetic energy of electrons ejected if radiation of wavelength 410 nm is used? We will use Equation 1.7 to find a) and then use Equation 1.6 to find the kinetic energy. 2.998 108 - -1 v =i=———X—m——s—=5.32x10‘4Hz 0 A0 564 X 10‘9 m ()5 = hi)O = (6.626 x 10'34 J-s) (5.32 x 1014 Hz) = 3.52 x 10-19 J hc KE=hv—¢=T—¢ _ (6.626 x 10-34 J-s) (2.998 X 108 m-s-') _ 410 x 10"9 m = 1.32 x 10-19 J — 3.52 X 10“9 J 1—22. A ground-state hydrogen atom absorbs a photon of light that has a wavelength of 97.2 nm. It then gives off a photon that has a wavelength of 486 nm. What is the final state of the hydrogen atom? First, we find the value of n2, the state of the hydrogen atom that is obtained upon absorption, by using Equation 1.10 with n] = l. 1 97.2 X 10’7 cm I1,=4 = 109 680 (1 — cm" "2 We can now use Equation 1.10 with r11 = 4 to find the final state of the hydrogen atom: —1 1 1 .—"—‘“9;2 m4 = 109 680 — —2) cm" é/gwmcm " ,FZ "2 A—3 , a. 6i=rc036+irsin9 r=6 ét‘l—tan‘1 6—) ‘ 0 0:1 2 so 6i 2 66"” b. 4-~/§i=rc030+irsin0 r=«/16+2=3«/§ — 2 9 = tan" = —0.340 SO 4 — fit = 3J5 e-°~34°" c. —1—2i=rcos0+irsino r=m=d§ —2 6 = tan‘1 = 1.11 so _1_2i = fielJli d. n+ei=rcos6+irsin6 r=\/712+e2 9 =tan‘1 (3) =0.7130 71' so 7: + ei = Vnz + e2 eo‘mi A—6. Show that eifi +e—i9 cos6 = — 2 and that eis __ e—ia sine = 2i Using Equation A.6, e” = c036 + i sine e‘” = cos0 — i sin6 Adding these two expressions gives e” + e‘”) = 2cos€ eia +e-i9 = 6 2 cos and subtracting the first two expressions gives em — e‘” = 2i sinO eta _ e—ie 2i = sin 6 A—9. Consider the set of functions <1>m(¢) = em m = 0, i1, :l:2, l 427: 05¢527r First show that 271 / d¢<I>m(¢) = O for all value of m :,£ 0 o . i 2 V2” m = 0 Now show that 27! d¢<I>:;(¢)<I>n(¢) = 0 m 7! n 0 =1 m=n 1 M11) ¢m(¢)=me m=0,:‘:1,j:2,... 0591,52,, Letm=O.Then 2" 1 2H —d =——=m l0 v27: ¢ «271 Ifm=,£0, 2n 1 I 2n 1 2n zm¢ _ ' /0 me dqb ——/0 m cos m¢d¢ +-/(; J27! s1nm¢d¢ (1) Each integral in Equation 1 is equal to zero, because they are evaluated over one full cycle of the function. Now consider 2:! 1 27! I ' / d¢¢;.(¢)‘p,.(¢) = — / d¢e""’¢e'"¢ o 277 o 1 271’ = _ i(n—m)¢ 27f 0 d¢e If n 7! m, we can define n — m = k and the resulting integral is identical to that in Equation 1, and so has a value of 0. If n = m, 2]! 1 27f / d¢<1>;,(¢)d>m(¢)=2— f d¢=1 0 TI 0 2—8. Consider the linear second-order differential equation dzy dy — + a x — + a x x = O dxz 1( )dx 0( )y() Note that this equation is linear because y(x) and its derivatives appear only to the first power and there are no cross terms. It does not have constant coefficients, however, and there is no general, simple method for solving it like there is if the coefficients were constants. In fact, each equation of this type must be treated more or less individually. Nevertheless, because it is linear, we must have that if y1(x) and y2(x) are any two solutions, then a linear combination, y(x) = clyl(x) + czyz(x) where c1 and c2 are constants, is also a solution. Prove that y(x) is a solution. Let y(x) = cly1(x) + c2y2(x). Then dy _ dy1 dyz dx _cldx +czdx d2 d2 d2 —Z2 = 61—yzl + 62—y22 dx dx dx d2 d d2 d2 d d —y +a1(x)fi + a0(x)y(x) = cl—Zl + c i +a1 [c A + ch—iz] +ao(c1y1+ czy d):2 dx2 2 dx2 1 dx dzy dy dzy dy Cl “I? New] + 62 + Hob] = c1(0) + c2(0) = 0 If yl(x) and y2(x) are solutions, y(‘x) = c1y1(x) + c2y2(x) is also a solution. 2—9. We will see in Chapter 3 that the Schrodinger equation for a particle of mass m that is constrz to move freely along a line between 0 and a is (12¢ 8n2mE 3% h2 )‘Wfl’ with the boundary condition W0) = Il/(a) = 0 In this equation, E is the energy of the particle and 1/1(x) is its wave function. Solve this differential equation for 1/}(x), apply the boundary conditions, and show that the energy can have only the values ’ nzh2 En=8 2 n=l,2,3,... ma or that the energy is quantized. Using the solution‘found in Example 2—4, with (02 = 8712mE/ hz, we find 1/; = Csinkx + Dcoskx where k = (87r2m E / h2)1/2. Applying the boundary condition 1M0) = 0 gives 1/; = C sin kx. The boundary condition tfla) = C sin ka = 0 requires that ka = mt with n = 1, 2, . . .. Therefore w(x)=Csini’l;1 n=l,2,3,... and ka=mt OI' Solvingrfor E gives 2h2 En = n 2 8ma n=1,2,3,... where we have added the subscript “n” to E to explicitly show the dependence of E on the integer n. ...
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KeyHW1 - 1—7. Sirius, one of the hottest known stars, has...

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