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Unformatted text preview: Chapter 12: From Classical to Quantum Mechanics Problem numbers in italics indicate that the solution is included in the Student ’5 Solutions
Manual. Questions on Concepts Q12.1) Why is there an upper limit to the photon energy that can be observed in the emission spectrum
of the hydrogen atom? The energy levels of the H atom are quantized and the largest energy corresponds to n —>oo. The highest
photon energy corresponds to the energy released when an electron makes a transition to the n = 1 state.
Q12.2) Why were investigations at the atomic and subatomic levels required to detect the wave nature
of particles? The wavelength of particles with greater mass is so short that particle diffraction could not be observed
experimentally. Therefore, the wave—particle duality present even in heavier particles was not observed.
Q12.3) Classical physics predicts that there is no stable orbit for an electron moving around a proton.
What criterion did Niels Bohr use to deﬁne special orbits that he assumed were stable? Bohr assumed that there were an integral number of wavelengths in the orbit. Q12.4) You observe light passing through a slit of width a and transition from the range 1 >> a to xi << a. Will you observe a sharp transition between ray optics and diffraction? Explain why or why not.
No. The transition occurs continuously with subsidiary diffraction peaks moving closer and closer to the
central diffraction peak. When we say that diffraction does not occur, we mean that it cannot be
experimentally observed. (212.5) Which of the experimental results for the photoelectric effect suggests that light can display
particle—like behavior? The fact that at very low intensity the incident light is still able to eject electrons from a material if its
frequency is above the threshold frequency suggests that the energy of the light can be concentrated in a
region of atomic dimensions at the surface of the material. This suggests particle—like behavior. Q12.6) Is the intensity observed from the diffraction experiment depicted in Figure 12.6 the same for
the angles shown in parts (b) and (c)? Yes. For all minima in a graph of intensity versus frequency, the intensity is zero. as seen in Figure
12.5. Q12.7) What feature of the distribution depicted as Case 1 in Figure 12.7 tells you that the broad distribution arises from diffraction? 271 Chapter 12/ From Classical to Quantum Mechanics The intensity goes through a minimum and increases again on both sides of the maximum. This can
only occur through wave interference. (212.8) Why does the analysis of the photoelectric effect based on classical physics predict that the
kinetic energy of electrons will increase with increasing light intensity? In the classical theory, more light intensity leads to more energy absorption by the electrons at the
surface of the solid. At equilibrium, as much energy must leave the surface as is absorbed. Therefore,
the electrons will have a greater energy. Classical physicists thought that the greater energy would
manifest as a higher speed. In fact, it appears as a larger number of electrons leaving the surface with a
frequency determined by the photon energy and the work function of the surface. Q12.9) In the double—slit experiment, researchers found that an equal amount of energy passes through
each slit. Does this result allow you to distinguish between purely particle—like and purely wavelike
behavior? No. This result would be expected for both waves and particles. In the case of particles, the same
number would pass through each slit for a large number of incident particles. For a wave, the fraction of
the intensity of the wave that passes through each slit would be the same. Q12.10) The inability of classical theory to explain the spectral density distribution of a blackbody was
called the ultraviolet catastrophe. Why is this name appropriate? The divergence between the classical theory and the experimental results is very pronounced in the UV region. In particular, because the classical theory predicts that the spectral density increases as v2 , the
total energy radiated by a blackbody is predicted to be inﬁnite. This prediction was clearly wrong, and
the lack of agreement with the experimental results was viewed by scientists of the time as a catastrophe.
Q12.11) In the diffraction of electrons by crystals, the volume sampled by the diffracting electrons is on
the order of 3 to 10 atomic layers. If He atoms are incident on the surface, only the topmost atomic layer
is sampled. Can you explain this difference? Whereas electrons can penetrate through the topmost layer into the solid, He atoms are too large to allow
penetration at thermal energies. Therefore electrons sample a number of atomic layers below the
surface, and He atoms are only sensitive to the outermost layer. Q12.12) Why is a diffraction pattern generated by an electron gun formed by electrons interfering with
themselves rather than with one another? All particles of a given wavelength have a random phase difference with another particle. The net effect
of adding the amplitudes of many different particles is that they cancel one another out. Therefore no diffraction is observed. 272 Chapter 12/ From Classical to Quantum Mechanics Q12.13) What did Einstein postulate to explain that the kinetic energy of the emitted electrons in the
photoelectric effect depends on the frequency? How does this postulate differ from the predictions of
classical physics? Einstein postulated that the energy of light depends on the frequency, whereas in classical theory, the
energy depends only on the intensity and is independent of the frequency. He further postulated that the
energy of light could only be an integral multiple of hv. Q12.14) How did Planck conclude that the discrepancy between experiments and classical theory for
blackbody radiation was at high and not low frequencies? He looked for a range of frequencies in which the classical theory and the experimental results agreed,
which is in the limit of low frequencies. Divergence is observed at high frequencies. Q12.15) Write down formulas relating the wave number with the frequency, wavelength, and energy of a photon. N V 1 E
V:—:— ——
c [t he Problems P12.1) When a molecule absorbs a photon, both the energy and momentum are conserved. If a H2 molecule at 750. K absorbs an ultraviolet photon of wavelength 225 nm, what is the change in its
velocity Av ? Given that its average speed is vm, = ,[3kT/ m , what is A v/vrm ? Because momentum is conserved, 434
= ﬂ zw = 2.94x10—27 kg m s—1 pphoton /1 225 X 1 0‘9 m All this momentum is transferred to the H2 molecule ApH2 = 2.94x10‘27 kg m s‘1 = mAv
2.94x10‘27 kg m 5“ 2.016 amuxl.66lx10“27 kg/amu 1 —1
Av : 0.879 ms = 0.879 ms 2 2.89x10‘4 v (3RT 3x8.3l4Jmol" K" x298K
M 2.016x10’3 kgmol“l AVHZ 2 = 0.879 m s“ 273 Chapter 12/ From Classical to Quantum Mechanics P12.2) A more accurate expression for E... would be obtained by including additional terms in the TayloriMaclaurin series. The Taylor—Maclaurin series expansion of ﬁx) in the vicinity ofxo is given by (see Math Supplement)
+i{————d3f 9)) (x~x0)3 +
3! d x H0 [21/
Use this formalism to better approximate E m by expanding e” in powers of hv/kT out to (fw/kT)3 in the vicinity of hV/kT = 0. Calculate the relative error, M, if you had not included the additional E().\C
terms for v: 1.00 X 101:2 s'1 at temperatures of 800., 500., and 250. K. Explain the trend you see. hi! 2 3
The Taylor series expansion of e“ is l + E + l [ 111/] + l [ hy] + . Therefore including terms up kT 2 k_T 6 If
hv 3
mi kT)’
— hv
E(ISL':——§—__——3‘
hv l[hv] l[hv]
kT 2 kT 6 kT
7,—kT I h
().\(,_ : ’lV2 5 [CT ~—‘—.V2_~ a?
E... hv 1(hv] 1(hv] hv 1(a) 1[hv)‘
—+#— +—— —+—— +——
kT 2kT 6kT kT 2kT 6kT
Elm—kTA
E. A 6626><IO “ Jsxmon” s' l[6.626x10 "4 J sxmoxlo'2 s ‘12 +
1.381x10 3‘ JK l><800K 2 l.381x10 23 J K ‘x800K H I
el,381xl0“'JK XSOOK x 6.626x10”Js><l.00x10'3s' J z
‘ {ammo “.1 sxl.00x10'zs ‘J' + 2} r] r
6 l.381><l() Jlx x8001\ 6.626x10 "4 JsxlOOon” s‘ 1(6626x10 34 .stIooXm” s 'I ’) + ’7
6626 10“] 100 10,. l l.38><lO"‘JK ‘xsoox 2 l.38><10“"]l< 'x800K
. X SX X M S +1 6.626x10 fj‘ .lsxl.00x10'2 s‘ x
6 1.381x103‘JK 'x800K E _
—i“—E_—£ = —0.0306 for 800. K. The corresponding values for 500. K and 250. _K are —0.0495 036 274 Chapter 12/ From Classical to Quantum Mechanics and ~0. 102. We expect the high temperature limit of Em, to be kT. Therefore the error is small at 800. K but becomes larger as T decreases. P12.3) The observed lines in the emission spectrum of atomic hydrogen are given by ~ 1 ,
V(cm"‘) 2 RH (cm"1) {j—izj cm 1, n > 111
n; n
In the notation favored by spectroscopists, 17 21/2 = E/hc and RH 2109,737cm“1. The Lyman, Balmer, and Paschen series refers to m = 1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of 17 and E in each of these series? The highest value for 17 corresponds to l —> 0. Therefore n 17 11 RH [T15] em"1 = 109,677 cm’lor Emax = 2.17871 x 10'18 J for the Lyman series. 17 2 RH [2%) em“1 = 27419.3 cm‘1 or Emax = 5.44676 x 10—19 J for the Balmer series, and I7 2 RH [317] cm71 2 12186.3 cm‘1 or Emax = 2.42078 x 10‘19 J for the Paschen series P12.4) Calculate the speed that a gas—phase nitrogen molecule would have if it had the same energy as
an infrared photon (/1 = 1.00 x 104 nm), a Visible photon (/1 : 500. nm), an ultraviolet photon (A =
100. nm), and an X—ray photon (/1 = 0.100 nm). What temperature would the gas have if it had the same
= (v2)l/2 : 3kT/m , for this energy as each of these photons? Use the root mean square speed, v rm A‘ calculation. V_ [25 _ ’2hc _ 2x 6.626x10'34J sx2.998><108m s"
m m1 28.02 amux1.661><10‘27 kg (amu)"xi.oox104x10*" m = 924 m s‘1 for 2 =1.00 x104 nm.
The results for 500. nm, 100. nm and 0.100 nm are 4.13 >< 103 m s“, 9.24x 103 m s‘l, and
2.92 x 105 m s“. 275 Chapter 12/ From Classical to Quantum Mechanics We calculate the temperature using the formula 2
4 28.02x10‘3kr mol‘lx 924m s"'
T_ A VIII“ 2 & —l( —' ) :959 K
3R 3x8.3l4Jmol K For it = 1.00 x 104 nm. The results for 500nm, 100. nm and 0.100 nm are 1.92 x 104 K, 9.59x 104 K,
and 9.59 ><107 K. PI2.5) Calculate the highest possible energy of a photon that can be observed in the emission spectrum
of H. The highest energy photon corresponds to a transition from n I 00 to n 2 l. — 109677 (1_—1—]— — 109677 cm 1 00
E = hot? = 2.17871x10’18J
P12.6) What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.40 eV absorbs 7.75 x 10’3 J of radiation at a wavelength of 275 nm? What is the kinetic energy and velocity of the electrons emitted? A34 8 #1 —19
Ezhc ¢26.626x10 Jsx2398x10 ms —2.40er1'602X10 J
/l 275x10 m eV
=3.38><10‘19 J 19
VZ‘iZTf: 2X3 ——'3——8X121 J28.61x105ms“‘
9.109x10 kg n__E10/al _ Elolal _ 7.75X1073J EM... lac/t (6.626x1034 J sx2.998x108 m s")/275><10“9 m
=1.17x1016
electrons. P12.7) Show that the energy density radiated by a blackbody Ema) 87rhv3 1 V : LOO p(V’T)dV : L00 63 e/Iv/kT dv —1 depends on the temperature as 74. (Hint: Make the substitution of variables x 2 hV/kT .) The deﬁnite integral f [x3 /(e)r ~ 1)]dx = 72'4/15 . Using your result, calculate the energy density radiated by a blackbody at 1350. K and 5250. K. 276 Chapter 12/ From Classical to Quantum Mechanics 00 3
EM” 2 87rhv 1 dv. Let x =hl/ ' dx =—h—dv V 0 c3 e%_l kT’ kT
”gm/3 1 87:15?” °° x3 871?le4
I _3 7.0,, dV= 33 Ix 61562 3,3
0 c ek1_1 hc 0e—l lShc
5 4 4 5 —23 — 4 4
At1350.K, EV’ 3:222? = ”(1381“34 J? ) “1358012 3 =2.51><1(r3 Jm’3At5250. K,
C 15x(6.626><10‘ Js) (2.998x10 ms )
Em, _ 87z5x(1.381x1023 JK")4 ><(5250. K)4 = 0.575 J m‘3 V ‘15x(6.626x10‘34 J s)3x(2.998x108 ms’l)3 P12. 8) What speed does a N2 molecule have if it has the same momentum as a photon of wavelength 180. nm? h p:_:mH2VH2 x1 —34
V _ h 6.626x10 13 20.07911“ 3.] H2 _ ”1112/1 T 28.02 amu><l.661><10’27 kg(amu)il ><180.><10’9 m P12.9) A newly developed substance that emits 315 W of photons with a wavelength of 275 nm is
mounted in a small rocket initially at rest in outer space such that all of the radiation is released in the
same direction. Because momentum is conserved, the rocket will be accelerated in the opposite
direction. If the total mass of the rocket is 6.75 kg, how fast will it be traveling at the end of 365 days in
the absence of frictional forces? The number of photons is given by _[ ,1
E WattsleS 315wxIJS
n': W = W = 44 W 8 1 =4.36><102°s‘1
Emma" [15; 6.626x10 Js><2.998><10 m s
,1 275x10” m The momentum of one photon is _£_ 6.626x10‘34 J s A 275 10_9 =2.41x10‘27kg m s’1
x m P The force is given by the rate of change of momentum. d(n'p) 62’! F: =2.41x10—27 kgms“ x4.36x102°s_] =1.05x10‘6 kgms“2 277 Chapter 12/ From Classical to Quantum Mechanics The ﬁnal speed is given by I? ‘ —6 —2
V : v0 + at = —t = lgid—O—k—gE§—~x365daysx 86400 S
m 6.75 kg day 24.91 m s" P12.10) In our discussion of blackbody radiation, the average energy of an oscillator E = hV/(th/H —l) was approximated as EMC 2 hv/ [(1 + hV/kT) — 1] = kT for hv/kT << 1. Calculate the ().\C relative error = (E — E ) E in making this a proximation for V = 4.00x 1012 5—1 at temperatures of
P approx 6000., 2000., and 500. K. Can you predict what the sign of the relative error will be without a detailed calculation? hv exp[ﬂ]—1 E Relative Error : hV ex [ﬂj—l
p kT hv
ex {25]4
p kT 6.626 ><10_3’4 J isx4.00x10l2 s*1 M 1.381x10‘23 J K" x6000K
exp[6.626x10" Jsx4.00x10'2s*‘] 1 1.381><10'23 J K'1 x6000K 6.626x10'34 J s><4.00x1012 s"
[66266044 J sx4.00><1012 s']
exp l 1.381x10“23 JK“x6000K
E—E Relative Error =—E‘”)ﬂ = «0.0162 for T: 6000. K. The results for 2000. K, and 500. K are —0.0496 and —0.219. 278 Chapter 12/ From Classical to Quantum Mechanics E _ Eappmx  Because Eappmx = kT> E, the relative error = E IS always a negative number. P12.11) Using the root mean square speed, vm“, = (v2)1/2 : J3kT/ m , calculate the gas temperatures of H and Xe for which A = 0.35 nm, a typical value needed to resolve diffraction from the surface of a metal
crystal. On the basis of your result, explain why Xe atomic beams are not suitable for atomic diffraction experiments. (6.626x10“34 Js)2 3x1.381x10‘23 JK"><1.008 amux1.661x10“27 kg amu" ><(0.35><10"9 m)
=52K For H, = 7 For Xe, T: 0.40 K. At 0.40 K, Xe temperature is well below its liquefaction temperature. It will not be possible to make an
atomic beam of Xe atoms with this wavelength with conventional means. P12.12) Electrons have been used to determine molecular structure by diffraction. Calculate the speed of an electron for which the wavelength is equal to a typical bond length, namely, 0.175 nm. V_ p h 6.626x10‘34J s __:_=——_m——£=4.16x106ms‘1
m mil 9.109><10 kg><0.175><10 m P12.13) For a monatomic gas, one measure of the “average speed” of the atoms is the root mean square
speed, Vm“. = (V2>l/2 = «BkT / m , in which m is the molecular mass and k is the Boltzmann constant. Using this formula, calculate the de Broglie wavelength for H and Ne atoms at 250. and at 750. K. h g h _ 6.626x10‘34ls
mvms J3kTm 3x1.381x10*23iKc'x250.1<x1.008amux1.661x10*27kg amu" 21.59x10‘10m A: or H at 750. K. A: 9.19x10511m for He. For Ne, ,1 : 3.56x10‘11m and 2.05 x1041 m at 250. K and
750. K, respectively. P12.14) The distribution in wavelengths of the light emitted from a radiating blackbody is a sensitive
function of the temperature. This dependence is used to measure the temperature of hot objects, without making physical contact with those objects, in a technique called optical pyrometry. In the limit 2 hc/SkT . At what max (hc/xlkT) >> 1, the maximum in a plot of p(/l,T) versus xi is given by /l
wavelength does the maximum in p(/l,T) occur for T: 850., 1300., and 5500. K? 279 Chapter 12/ From Classical to Quantum Mechanics According to Example Problem 141, 21m : 5:67, . A ~ 6.626x10‘34 J sx2.998><108m s:l max 5 1381 10—23 “(in 850 K =3.38><10“°mat 850. K. AM for 1300. Kand 5500. Kis
X . X x . 2.21 x 10‘6 m and 5.23 x 10'7 m, respectively. P12.15) A beam of electrons with a speed of 4.75 x 104 m/s is incident on a slit of width 235 nm. The
distance to the detector plane is chosen such that the distance between the central maximum of the
diffraction pattern and the ﬁrst diffraction minimum is 0.375 cm. How far is the detector plane from the
slit? lhe dlffractlon mlnima satisfy the condrtlon sm 6 = —, n 2 i1, i 2, and the ﬁrst minimum IS at a sin 6 = ii. We choose the plus Sign (the minus sign gives the distance from the slit in the opposite a i h 6.626x10‘34 J s = 0.06517 . . . . Sing: 2 —3 4 41 79
d1rect1on)g1v1ng a mva 9.109x10 kgx4.75><10 ms x235><10 m «9 = 5.74 degrees
The distance d from the screen and the position of the ﬁrst minimum 5 are related by _ 5 _0.375 cm
tanH 0.0652 P1216) If an electron passes through an electrical potential difference of 1 V, it has an energy of 1 d 25.74 cm electronvolt. What potential difference must it pass through in order to have a wavelength of 0.225 nm? 2 7
Ezlm)v2=lnzcx[ h 1 h 2 t 2 my]. 22mg2
(6.626x10’34 J s)2 leV =———e———7x—~——;—=29.7 eV
2><9.109><10’3l kgx(2.25x10*'° my 1.602x10 '9 J The electron must pass through an electrical potential of 29.7 V.
P12.17) Calculate the longest and the shortest wavelength observed in the Balmer series. For the longest wavelength, the transition is from n = 3 to n = 2. 17 =109677[i——1—] = 15232 cm‘l 22 32 a = = 6.565x10“5 cm i
a 280 Chapter 12/ From Classical to Quantum Mechanics For the shortest wavelength, the transition is from n = 00 to n = 2. 17 =109677[§12——i2] = 27419 cm‘1
(X)
xi :—1= 3.647x10’5cm
v P12.18) X—rays can be generated by accelerating electrons in a vacuum and letting them impact on atoms in a metal surface. If the 1525—eV kinetic energy of the electrons is completely converted to the
photon energy, what is the wavelength of the X—rays produced? If the electron current is 2.25 x 10‘5 A, how many photons are produced per second? he _ 6.626x10‘34 J sx2.998x108 m s" ’12 E 1602 10*” J 208” mm
1525 er;
eV
75 A1
11 current 2 2.25x10 C s :1.40><1014 so]. _ charge per electron 1.602 ><10"9 C P12.19) The following data were observed in an experiment on the photoelectric effect from potassium: 10f9 Kinetic
4.49 3.09 1.89 1.34 0.700 0.311
0. 350. 400. 450. 500. Energy (J)
Graphically evaluate these data to obtain values for the work function and Planck’s constant. Wavelength 3.5 x135 3.9x 1&5 3.5 x1136 4.9x m‘imnbda l The best—ﬁt line is given by EU) 2 —3.95831x10‘19 + 2.1032><10'25 l 281 Chapter 12/ From Classical to Quantum Mechanics 2.11171x10’25 Jm 8 ‘l = 7.02x10“34J s. The work function is given by the
2.998X10 m S Because the slope is he, h = , ~25
intercept of the line with the x axis at y = 0. ¢ = E — E where A 0 = W =5.31x10*7 m
A, 3.97362X10“° J This gives ¢ = 3.74x107'9J or 2.33eV. P12.20) The power (energy per unit time) radiated by a blackbody per unit area of surface expressed in
units ofW m’2 is given by P 2 074 with o = 5.67 X 10’8 W m“2 K4. The radius of the sun is 6.95 x 105
km and the surface temperature is 5750. K. Calculate the total energy radiated per second by the sun.
Assume ideal blackbody behavior. E = PA: 0T4 x47rr2 = 5.67 x1078 w m’2K74 x (5750. K)4 x47r>< (6.95 x108 m)2 = 3.76 ><1026 w P1221) The work function of tungsten is 4.50 eV. What is the minimum frequency of light required to
observe the photoelectric effect on W? If light with a 225—nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons? a) For electrons to be emitted, the photon energy must be greater than the work function of the surface. 49
E =hv2450 erw=9.05xio'9 J
eV
719
v 2£>m2 1.09x1015 s"1 h * 6.626x10‘34 J s
b) The outgoing electron must ﬁrst surmount the barrier arising from the work function, so not all the photon energy is converted to kinetic energy. hc
E,=hv— 2——
. 75 1 ¢ _ 6.626x10'34 J s><2.998><108 m s“1
225x10"9 m E . "9
V, 2 c = 2X16”? J ~5.96x105 ms'l
my 9.11x10‘ kg P1222) Assume that water absorbs light of wavelength 3.50 x 10'6 m with 100% efﬁciency. How many 7.21x10"9 J=1.62x10"9 J photons are required to heat 2.50 g of water by 1.00 K? The heat capacity of water is 75.3 J mol’1 [(4. 282 Chapter 12/ From Classical to Quantum Mechanics E=th=N%=nC AT [1 ,m N— m CWAT/l _ 2.50 g 75.3 JK‘l mol‘1x1.00Kx3.50x10’6 m M hc ‘ 18.02 g mor‘ 6.626x10‘34 Jsx2.998x108 m s“1
21.84x1020 P12.23) Calculate the longest and the shortest wavelength observed in the Lyman series. For the longest wavelength, the transition is from n = 2 to n = l. 17 = 109677 Gail—l = 82257.8 cm"
1=i~=121569x1075cm
V For the shortest wavelength, the transition is from n = 00 to n = 1. a = 109677[l—1—2] = 109677 cm‘1 1 oo 71:1—911768x10‘6cm V P12.24) A 1000—W gas discharge lamp emits 5.25 W of ultraviolet radiation in a narrow range centered near 325 nm. How many photons of this wavelength are emitted per second? I Emm, 3.00WXlJ 3’] W‘1 5.25WX] J 3‘1 W71 18 7.]
n 2 :———:_—‘_34——_‘T‘_7T=8.59x10 3
EM” LC 6.626x10 Js x2.998><10 m s
’1 325x109 m P12.25) The power per unit area emitted by a blackbody is given
by P 2 CT" with 0' = 5.67><10“8 W III—2 K‘4 . Calculate the energy radiated per second by a spherical blackbody of radius 0.325 m at 850. K. What would the radius of a blackbody at 3250. K be if it emitted
the same energy as the spherical blackbody of radius 0.325 m at 850. K? E = A0T4 :47I(O.325m)2 x5.67><10’8J s" 11172 K’4 ><(850. K)“ = 3.93x104J 5’1 Because the total energy radiated by the spheres must be equal, 2 4 2 4
47H"1 O'Tl =4irr2 O'T2 r : [712714 = (0.325m)2(850.K)4 20022 m
2 T2“ (3250. K)“ P12.26) A ground state H atom absorbs a photon and makes a transition to the n = 4 energy level. It then emits a photon of energy 4.085 x 10’19 J. What is the ﬁnal energy and n value of the atom? The ﬁrst four energy levels of the H atom are given by 283 Chapter 12/ From Classical to Quantum Mechanics E :_2.179><10"8 J n2
E1:—2.179><10‘18 J
E2 =—5.448x10*"’ J
E3 =—2.421x10‘19 J
E =—1.362x10"9 J E4 —4.085x10*l9 J=—1.362x10"" J—4.085><10“9 J=—5.447><10"° J This energy corresponds to the n I 2 state of H. P12.27) Pulsed lasers are powerful sources of nearly monochromatic radiation. Lasers that emit photons in a pulse of lOns duration with a total energy in the pulse of 0.10 J at 1000 nm are commercially available. a. What is the average power (energy per unit time) in units of watts (l W = l J/s) associated with such
a pulse? b. How many 1000hm photons are emitted in such a pulse? a) P;é£:QL_1.%:1,0X107JS_I
Al 1.0X10 S Epulse : EPU/SC : “L 1‘ 5.0X1017 8 ‘1
EW’" hE 6.626x10’34 ”43—92%
A 1.000x10 m b)N= 284 ...
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