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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 103 mK 2.90 x 10‘3 mK 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 Js)(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'”Jphoton‘ 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 Js) (5.32 x 1014 Hz) = 3.52 x 1019 J
hc KE=hv—¢=T—¢ _ (6.626 x 1034 Js) (2.998 X 108 ms')
_ 410 x 10"9 m
= 1.32 x 1019 J — 3.52 X 10“9 J 1—22. A groundstate 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 +ei9 = 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 secondorder 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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 Fall '09
 Mccurdy
 Energy, EIA

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