UNIST

20131241 박용현

31. 전자기 진동과 교류

LC 진동

𝐶𝐿

𝑖

𝑖

𝑈 𝐵 𝑈𝐶

𝑈 𝐵=12 𝐿 𝑖2

𝑈𝐶=12𝐶 𝑞2

LC 진동

𝐶𝐿

𝑖

𝑖

𝑈 𝐵 𝑈𝐶

𝑈 𝐵=12 𝐿 𝑖2

𝑈𝐶=12𝐶 𝑞2

LC 진동

𝐶𝐿

𝑖

𝑖

𝑈 𝐵 𝑈𝐶

𝑈 𝐵=12 𝐿 𝑖2

𝑈𝐶=12𝐶 𝑞2

LC 진동

𝐶𝐿

𝑖

𝑖

𝑈 𝐵 𝑈𝐶

𝑈 𝐵=12 𝐿 𝑖2

𝑈𝐶=12𝐶 𝑞2

LC 진동

𝐾=12𝑚𝑣2

𝑀 𝑈 𝐾

𝑈=12 𝑘𝑥

2

LC 진동𝐾=

12𝑚𝑣2

𝑀𝑈=

12 𝑘𝑥

2

𝐶𝐿

𝑖

𝑖

𝑈𝐶=12𝐶 𝑞2 𝑈 𝐵=

12 𝐿 𝑖2

𝑣=𝑑𝑥𝑑𝑡

𝑖=𝑑𝑞𝑑𝑡

𝑥⟷𝑞𝑇=2𝜋 √𝑚𝑘 ⟷2𝜋 √𝐿𝐶

LC 진동

𝑀

𝐶𝐿

𝑖

𝑖

𝑈 𝑡𝑜𝑡=𝑈+𝐾=12 𝑘𝑥

2+12𝑚𝑣2=𝑐𝑜𝑛𝑠𝑡

𝑑𝑈 𝑡𝑜𝑡

𝑑𝑡 =12𝑘𝑥 𝑑𝑥

𝑑𝑡 + 12𝑚𝑣 𝑑𝑣

𝑑𝑡 =0

𝑈 𝑡𝑜𝑡=𝑈𝐶+𝑈 𝐵=12𝐶 𝑞2+ 12 𝐿𝑖

2=𝑐𝑜𝑛𝑠𝑡

𝑑𝑈 𝑡𝑜𝑡

𝑑𝑡 = 12𝐶 𝑞 𝑑𝑞

𝑑𝑡 + 12𝐿𝑖 𝑑𝑖𝑑𝑡=0

교류 회로 ( 저항 )

𝑡

𝑉𝑉=𝑉𝑚 sin (𝜔𝑡 ) 𝑉 𝑅

교류 회로 ( 축전기 )

𝑡

𝑉 𝐶

𝑖=𝑉𝑅⟷𝑖=𝜔𝐶𝑉 ⇒ 𝑋𝐶=

1𝜔𝐶

𝑉𝑉=𝑉𝑚 sin (𝜔𝑡 )

교류 회로 ( 유도기 )

𝑡

𝑉𝑉=𝑉𝑚 sin (𝜔𝑡 )

𝑉 𝐿

𝑖=𝑉𝑅⟷𝑖= 𝑉

𝜔𝐿⇒ 𝑋 𝐿=𝜔𝐿

2. An ac generator with emf, where and , is con-nected to a 4.15 mF capacitor. (a)What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the gen-erator? (c)When the emf of the generator is -12.5 V and increasing in magnitude, what is the current?

𝑉𝑉=𝑉𝑚 sin (𝜔𝑡 )

𝑉 𝐶

𝑡

41. A variable capacitor with a range from 10 to 410 pF is used with a coil to form a variable-frequency LC circuit to tune the in-put to a radio. (a) What is the ratio of maximum frequency to min-imum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.54 MHz to 1.60 MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance should be added and (c) what inductance should the coil have?

𝐶𝑣𝑎𝑟𝐿

𝜔=1

√𝐿𝐶𝜔𝑚𝑎𝑥=

1√𝐿𝐶𝑚𝑖𝑛

𝜔𝑚𝑖𝑛=1

√𝐿𝐶𝑚𝑎𝑥𝐶

𝜔𝑚𝑎𝑥

𝜔𝑚𝑖𝑛=√𝐶𝑚𝑎𝑥

𝐶𝑚𝑖𝑛

√𝐶𝑚𝑎𝑥

𝐶𝑚𝑖𝑛=6.4

𝜔𝑚𝑎𝑥

𝜔𝑚𝑖𝑛=2.96

𝜔𝑚𝑎𝑥

𝜔𝑚𝑖𝑛=√𝐶𝑚𝑎𝑥+𝐶

𝐶𝑚𝑖𝑛+𝐶𝜔𝑚𝑎𝑥=

1√𝐿(𝐶¿¿𝑚𝑖𝑛+𝐶 )=2𝜋 𝑓 𝑚𝑎𝑥 ¿

44. An ac generator has , where and It is connected to a 12.7 H inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is -15 V and increasing in magnitude, what is the current?

𝑉𝑉=𝑉𝑚 sin (𝜔𝑡)

𝑉 𝐿

𝑡

57. An oscillating LC circuit consisting of a 1.0 nF ca-pacitorand a 9.0 mH coil has a maximum voltage of 3.0 V.What are (a) the maximum charge on the capacitor, (b) the maximum current through the circuit, and (c) the maximum energy stored in the magnetic field of the coil?𝑞=𝐶𝑉𝑚sin (𝜔𝑡 )=𝑞𝑚 sin (𝜔𝑡 )

𝑈 𝑡𝑜𝑡=𝑈𝐶+𝑈 𝐵=12𝐶 𝑞2+ 12 𝐿𝑖

2=12 𝐿𝑖𝑚

2=12𝐶 𝑞𝑚

2

𝑖=𝑑𝑞𝑑𝑡 =𝜔𝐶𝑉𝑚cos (𝜔𝑡 )

62. In an oscillating LC circuit, when 30.0% of the total energy is stored in the inductor’s magnetic field, (a) what multiple of the maximum charge is on the capac-itor and (b) what multiple of the maximum current is in the inductor?

𝑈𝐶=0.7𝑈 𝑡𝑜𝑡=0.712𝐶 𝑞𝑚

2=12𝐶 𝑞2

𝑈 𝐵=0.3𝑈 𝑡𝑜𝑡=0.312 𝐿𝑖𝑚

2=12 𝐿𝑖

2

64. The frequency of oscillation of a certain LC circuit is 200 kHz. At time , plate A of the capacitor has max-imum positive charge.At what earliest time will (a) plate A again have maximum positive charge, (b) the other plate of the capacitor have maximum positive charge, and (c) the inductor have maximum magnetic field?

𝐶𝐿

𝑖

𝑖

𝐴

64. The frequency of oscillation of a certain LC circuit is 200 kHz. At time , plate A of the capacitor has max-imum positive charge.At what earliest time will (a) plate A again have maximum positive charge, (b) the other plate of the capacitor have maximum positive charge, and (c) the inductor have maximum magnetic field?

𝐶𝐿

𝑖

𝑖(a)

𝐴

65. (a) At what frequency would a 12 mH inductor and a 10 mF capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same L and C.

𝑋𝐶=1

𝜔𝐶𝑋𝐿=𝜔𝐿

32. 맥스웰 방정식

전기장에서의 가우스 법칙

𝑄

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝑒𝑛𝑐

패러데이 법칙

∮𝐸 ∙𝑑 �⃗�=−𝑑Φ𝐵

𝑑𝑡

자기장에서의 가우스 법칙N

S

Φ𝐵=∫ �⃗� ∙𝑑 �⃗�=0

유도 자기장 ( 앙페어 - 멕스웰 법칙 )

𝜀=−𝑑Φ𝐵

𝑑𝑡∮ �⃗� ∙𝑑 �⃗�=𝜀0𝜇0

𝑑Φ𝐸

𝑑𝑡

∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐+𝜀0𝜇0𝑑Φ𝐸

𝑑𝑡

∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐

변위 전류∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐+𝜀0𝜇0

𝑑Φ𝐸

𝑑𝑡

𝑖𝑒𝑛𝑐⟷𝜀0𝑑Φ𝐸

𝑑𝑡

−𝑞

+𝑞𝑑

𝑉𝐸

𝑖

𝑖𝑑=𝜀0𝑑Φ𝐸

𝑑𝑡

33. 전자기파

전자기파• 전자기파의 진행방향은 전기장과 자기장의 방향과 항상 수직• 전기장의 방향과 자기장의 방향은 수직 방향은 전자기파의 진행방향• 전기장과 자기장은 sin 파형 . 𝑐=

1√𝜀0𝜇0

=𝐸𝐵

Poynting vector• 방향 : 전자기파의 진행방향• 크기 : 전자기파의 세기와 관련됨

Intensity

𝑇

𝑋

𝑋 𝑇=∫0

𝑇

𝐸𝑚2𝑠𝑖𝑛2(𝑘𝑥−𝜔𝑡)  𝑑𝑡 𝐼=𝑆𝑎𝑣𝑔=

1𝑐𝜇0

𝐸𝑚2

2= 1𝑐𝜇0

𝐸 𝑟𝑚 𝑠2

편광

편광

𝐼 1=12 𝐼 0

𝐼 0

𝐼 1

편광

𝜃

𝐸1=𝐸0𝑐𝑜𝑠𝜃𝐼 ∝ 𝐸2

𝐼 1=𝐼 0𝑐𝑜𝑠2𝜃

반사와 굴절

𝜃1 𝜃1

𝜃2

𝑛1𝑛2

𝑛1 sin𝜃1=𝑛2 sin 𝜃2

반사와 굴절 ( 전반사 )

𝑛1𝑛2

𝑛1 sin 90=𝑛2 sin 𝜃𝐶 (𝑛2>𝑛1)

𝜃𝑐

브루스터의 법칙

𝜃1 𝜃1

𝜃2

𝑛1𝑛2

𝜃1=𝜃𝐵

34.images

평면 거울

𝑂 𝐼

𝑝 𝑖

𝑖=−𝑝

구형 거울

concave mirror( 오목 거울 )convex mirror( 볼록 거울 )

구형 거울

𝐶 𝐹

𝑓𝑟

𝑟=2 𝑓

구형 거울

𝐹 𝐶

𝑟𝑓

𝑟=2 𝑓

구형 거울

𝐶 𝐹

1𝑝+

1𝑖 =

1𝑓

구형 거울

𝐶 𝐹

1𝑝+

1𝑖 =

1𝑓

𝑓 <𝑝

𝑝< 𝑓

𝑖 𝑚=− 𝑖𝑝𝑅 /𝑉𝐼 /𝑁𝐼

구형 거울

𝐶 𝐹

1𝑝+

1𝑖 =

1𝑓

𝑓 <𝑝

𝑝< 𝑓

𝑖 𝑚=− 𝑖𝑝𝑅 /𝑉𝐼 /𝑁𝐼

𝑂

𝑝

+¿ − 𝑅 𝐼

𝑂

− +¿ 𝑉 𝑁 𝐼

구형 거울

𝐹 𝐶

− 𝑓 <𝑝

𝑝<− 𝑓

𝑖 𝑚=− 𝑖𝑝𝑅 /𝑉𝐼 /𝑁𝐼

− +¿ 𝑉 𝑁 𝐼

− +¿ 𝑉 𝑁 𝐼

1𝑝+

1𝑖 =

1𝑓

𝑓

𝑂

1𝑖 =

𝑝− 𝑓𝑝𝑓

𝑂

구형 굴절면𝑛1𝑝 +

𝑛2𝑖 =

𝑛2−𝑛1𝑟

𝑛1𝑛2

𝑂 𝐶

𝑟

concave −

convex +¿

𝑖<0

𝑅 /𝑉

𝑉

𝑖>0 𝑅

𝐼𝑖

얇은 렌즈1𝑓 =(𝑛−1)( 1𝑟1

− 1𝑟 2)

1𝑝+

1𝑖 =

1𝑓

𝑛 𝑛concave lens( 오목 렌즈 )

Convex lens( 볼록 렌즈 )

𝑟1 𝑟1𝐶1 𝐶1𝐶2 𝐶2

𝑟2 𝑟2

얇은 렌즈1𝑓 =(𝑛−1)( 1𝑟1

− 1𝑟 2)

1𝑝+

1𝑖 =

1𝑓

𝑛concave lens( 오목 렌즈 )

Convex lens( 볼록 렌즈 )

𝑟1 𝑟1

𝐶1 𝐶1𝐶2 𝐶2

𝑟2 𝑟2

𝑟

concave −

convex +¿

𝑛𝑟1 :− 𝑟1 :+¿

Two lens system

𝑂

1𝑝+

1𝑖 =

1𝑓

1𝑓 =(𝑛−1)( 1𝑟1

− 1𝑟 2)

𝑖𝑂 𝑖 ′

9. A lens is made of glass having an index of refraction of 1.5.One side of the lens is flat, and the other is con-vex with a radius of curvature of 20 cm. (a) Find the focal length of the lens. (b) If an object is placed 40 cm in front of the lens, where is the image?

𝐶2

𝑟2 𝑟1=∞

(𝑎)1𝑓 =(𝑛−1)( 1𝑟 1

− 1𝑟 2)

𝑟1 :+¿

(𝑏)1𝑝 +

1𝑖 =

1𝑓

10. a real inverted image I of an object O is formed by a certain lens (not shown); the object–image separa-tion is d =50.0 cm, measured along the central axis of the lens. The image is just half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens?

𝑂𝐼

𝑑𝑝 𝑖

𝑝+𝑖=𝑑

1𝑝+

1𝑖 =

1𝑓

1𝑓 =(𝑛−1)( 1𝑟1

− 1𝑟 2)

convexconcave

𝑟1 :−concave𝑟1 :+¿convex

12. An object is moved along the central axis of a thin lens while the lateral magnification is measured. ver-sus object distance out to .0cm to .0cm What is the m for.0?

𝑝𝑎 𝑝𝑏𝑝 (𝑐𝑚)

𝑚2

4

𝑚=2=− 𝑖𝑝

1𝑝+

1𝑖 =

1𝑓

47. a beam of parallel light rays from a laser is inci-dent on a solid transparent sphere of index of refrac-tion n. (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere? (b) What index of refraction, if any, will pro-duce a point image at the center of the sphere?

𝑖=2𝑟𝑝=∞

𝑛1𝑝 +

𝑛2𝑖 =

𝑛2−𝑛1𝑟

𝑖=𝑟

93. A concave shaving mirror has a radius of curvature of 35.0 cm. It is positioned so that the (upright) image of a man’s face is 1.70 times the size of the face. How far is the mirror from the face?

𝑚=1710=− 𝑖

𝑝1𝑝 +

1𝑖 =

1𝑓 =

2𝑟

𝐶 𝐹

35. 간섭

굴절의 법칙

𝑟1𝑟2

𝜃1 𝜃1𝜃1𝜃2 𝜃2

𝜃2

h 𝜆1

𝜆2

h sin 𝜃1=𝜆1

굴절의 법칙

𝑛1𝑛2

𝑟1𝑟2

𝜃1 𝜃1𝜃1𝜃2 𝜃2

𝜃2

h 𝜆1

𝜆2

h sin 𝜃1=𝜆1

회절

영의 간섭실험

영의 간섭실험 𝐷

𝑑 𝜃𝜃

𝑑sin 𝜃

강쇄

𝜃슬릿 1

슬릿 2

영의 간섭실험 𝐷

𝑑 𝜃𝜃

𝑑sin 𝜃

𝜃 𝑦 1

( 보강 )( 상쇄 )

얇은 필름에의한 간섭

𝑛1

𝑛1

𝑛2𝐿

𝑛1<𝑛2

쇄강

𝜆1

𝜆2

고정단 반사

자유단 반사

31. a broad beam of light of wavelength 683 nm is sent directly downward through the top plate of a pair of glass plates. The plates are 120 mm long, touch at the left end, and are separated by 48.0 m at the right end. The air between the plates acts as a thin film. How many bright fringes will be seen by an observer looking down through the top plate?

a=48.0 m

고정단 반사자유단 반사

b=120 mm 2𝐿=(𝑚+

12 )𝜆

2𝑎=(𝑚𝑚𝑎𝑥+12 )𝜆

32. Two rectangular glass plates () are in contact along one edge and are separated along the opposite edge. Light with a wavelength of 600 nm is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the sepa-rated edges is increased by 600 nm, how many dark fringes will there then be across the top plate?

𝐿

2𝐿=𝑚𝜆 ,𝑚=0,1,2⋯ 8 (9개 )

𝐿 ′

𝐿′=𝐿+600𝑛𝑚 ,2𝐿 ′=𝑚𝜆2𝐿=(𝑚+

12 )𝜆 ,𝑚=0,1,2⋯ 7 (8개 )

41. In a double-slit experiment, the distance between slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480 nm, and the other due to light of wavelength 600 nm.What is the separa-tion on the screen between the third-order bright fringes of the two interference patterns?

𝑦𝑚=𝑚𝐷𝜆𝑑

52. A thin film coats a thick glass plateWhite light is incident normal to the film. In the reflec-tions, fully destructive interference occurs at 600 nm and fully constructive interference at 700 nm. Calcu-late the thickness of the film.

𝐿

쇄

𝑛=1.25𝑛=1.50

고정단 반사고정단 반사

62. In a double-slit experiment, the fourth-order maximum for a wave-length of 450 nm occurs at an angle of = 90°. (a) What range of wave-lengths in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-or-der maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?

(𝑎 ) , (𝑏) ,(𝑐)𝑑sin 90𝑚 <𝜆<700

75. The reflection of perpendicularly incident white light by a soap film in air has an interference maxi-mum at 600 nm and a minimum at 450 nm, with no minimum in between. If for the film, what is the film thickness, assumed uniform?.

𝐿

쇄강

36. 회절

𝑎𝜃

𝜃𝑎 sin𝜃

단일 슬릿에 의한 회절

단일 슬릿에 의한 회절

𝜆

𝜆2

쇄

𝑟1𝑟2𝑟3

𝑟 4

𝑟5

𝑟6

단일 슬릿에 의한 회절

3𝜆2

𝜆

maxima

𝜆2

𝑟1𝑟2𝑟3

𝑟 4

𝑟5

𝑟6

Circular aperture 에 의한 회절

번째 상쇄간섭 )𝑎 sin𝜃=𝜆  

𝑑𝜃

분해능

)

1

2 21

𝑑 𝜃

Grating 에 의한 회절

Grating

강 )

𝑑

𝑑

𝑑

𝑑

𝑑

𝑑

𝑟1𝑟2𝑟3𝑟 4𝑟5

𝑑sin 𝜃

𝑑sin 𝜃

𝑑sin 𝜃𝑑sin 𝜃

11. A grating has 400 lines/mm. How many orders of the entire visible spectrum ( nm) can it produce in a diffraction experiment, in addition to the order?

강 )

𝑑=𝑚𝜆

13. Estimate the linear separation of two objects on Mars that can just be resolved under ideal conditions by an observer on Earth (a) using the naked eye and (b) using the 200 in. ( 5.1 m) Mount Palomar tele-scope. Use the following data: distance to Mars km, diameter of pupil 5.0 mm, wavelength of light 550 nm.

1

2

𝜃𝑅

)

𝐿=8.0×107𝑘𝑚

𝑥 𝑑

𝑥=𝐿𝜃𝑅

16. Visible light is incident perpendicularly on a grat-ing with 315 rulings/mm.What is the longest wave-length that can be seen in the fifth-order diffraction?

𝑑sin 𝜃=𝑚 𝜆

37. A slit 1.00 mm wide is illuminated by light of wave-length 589 nm.We see a diffraction pattern on a screen 3.00 m away.What is the distance between the first two diffraction minima on the same side of the central dif -fraction maximum?

𝑎 sin𝜃=𝑚𝜆  

𝑦𝑚=𝑚𝐷𝜆𝑎

47. The two headlights of an approaching automobile are 1.4 m apart. At what (a) angular separation and (b) maximum distance will the eye re-solve them? Assume that the pupil diameter is 5.0 mm, and use a wave-length of 550 nm for the light. Also assume that diffraction effects alone limit the resolution so that Rayleigh’s criterion can be applied.

1

2𝑥=1.4𝑚 𝜃𝑅

𝐿

𝑑

)𝑥=𝐿𝜃𝑅

51. The distance between the first and fifth minima of asingle-slit diffraction pattern is 0.35 mm with the screen 40 cm away from the slit, when light of wave-length 550 nm is used. (a) Find the slit width. (b) Cal-culate the angle of the first diffraction minimum.

𝑎 sin𝜃=𝑚𝜆  

, mm

38. 광자와 물질파

광자•빛의 입자성

광전 효과

h𝑓 =𝐾𝑚𝑎𝑥+Φ

𝐾𝑚𝑎𝑥

𝑓Φ

광전 효과

𝑉

𝑓Φ𝑒

h𝑓 =𝐾𝑚𝑎𝑥+Φ,𝐾𝑚𝑎𝑥=𝑒𝑉

𝑉

전자와 물질파

𝑝=h𝜆

슈뢰딩거 방정식• 전자의 파동성 해석• 뢰딩거 방정식• 률밀도함수

불확정성 원리

Δ𝑥 Δ𝑝 ≥ℏ

3. In an old-fashioned television set, electrons are ac-celerated through a potential difference of 25.0 kV. What is the de Broglie wavelength of such electrons? (Relativity is not needed.)

𝑒𝑉=12𝑚𝑣2 ,𝑝=

h𝜆

13. The uncertainty in the position of an electron along an x axis is given as 50 pm, which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component of this electron?

Δ𝑥 Δ𝑝 ≥ℏ

16. An electron and a photon each have a wavelength of 0.20 nm. What is the momentum (in kg m/s) of the (a) electron and (b) photon? What is the energy (in eV) of the (c) electron and (d) photon?

𝑝=h𝜆

𝐾=12𝑚𝑣2

𝐸=h𝑓 =h 𝑐𝜆

25. The wavelength of the yellow spectral emission line ofsodium is 590 nm. At what kinetic energy would an electron have that wavelength as its de Broglie wave-length?

𝑝=h𝜆

𝐾=12𝑚𝑣2

34. A helium–neon laser emits red light at wavelength 633 nm in abeam of diameter 3.5 mm and at an en-ergy-emission rate of 5.0mW. A detector in the beam’s path totally absorbs the beam. At what rate per unit area does the detector absorb photons?

𝑃 𝑙𝑎𝑠𝑒𝑟=𝐸𝑙𝑎𝑠𝑒𝑟

𝑡 =5𝑚𝑊

𝐸 h𝑝 𝑜𝑡𝑜𝑛=h𝑓 =h 𝑐𝜆

42. The wavelength associated with the cutoff fre-quency forsilver is 325 nm. Find the maximum kinetic energy of electrons ejected from a silver surface by ultraviolet light of wavelength 254 nm.

𝐾𝑚𝑎𝑥

𝑓Φ

𝑓 0

h𝑓 =𝐾𝑚𝑎𝑥+Φ⇒ h 𝑓 0=0+Φh𝑓 =𝐾𝑚𝑎𝑥+h 𝑓 0

43. At what rate does the Sun emit photons? For sim-plicity, assume that the Sun’s entire emission at the rate of 3.9 W is at the single wavelength of 550 nm.

𝑃𝑠𝑢𝑛𝑡=𝐸𝑠𝑢𝑛=𝑁 h𝑝 𝑜𝑡𝑜𝑛𝐸 h𝑝 𝑜𝑡𝑜𝑛

𝑟𝑎𝑡𝑒=𝑁 h𝑝 𝑜𝑡𝑜𝑛

𝑡

55. The stopping potential for electrons emitted from a surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. (a) What is this new wavelength? (b) What is the work function for the surface?𝑉

𝑓Φ𝑒

h𝑓 =𝑒𝑉 +Φ

h ( 𝑓 2− 𝑓 1 )=𝑒(𝑉 2−𝑉 1)

77. X rays with a wavelength of 71 pm are directed onto agold foil and eject tightly bound electrons from the gold atoms.The ejected electrons then move in circular paths of radius r in a region of uniform magnetic field . For the fastest of the ejected electrons, the product is equal to 1.88 Tm. Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.𝐹=𝐵𝑒𝑣 ,𝐹=

𝑚𝑣2𝑟

𝐾=12𝑚𝑣2

h𝑓 =𝐾𝑚𝑎𝑥+Φ

39. 물질파

무한포텐셜장벽에 갇혀진 전자

0 𝐿

𝐸

𝑥𝐿

2𝐿=𝑛𝜆=𝑛 h𝑝=𝑛 h√2𝑚𝐸

무한포텐셜장벽에 갇혀진 전자𝐸𝑛=(

h2

8𝑚𝐿2)𝑛2

𝐸1

𝐸2

𝐸3

h𝑓 =Δ 𝐸=𝐸𝑚−𝐸𝑛

수소 원자의 보어 모델

𝑟

𝐷

1

1

2

2

2𝐷⇔2𝜋𝑟

수소 원자의 보어 모델

𝑟

4. An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 375 nm and then im-mediately emits a photon whose associated wave-length is 580 nm. How much net energy is absorbed by the atom in this process?

𝐸=h 𝑓 𝑎𝑏𝑠𝑜𝑟𝑏−h 𝑓 𝑒𝑚𝑖𝑡

48. An electron, trapped in a one-dimensional infinite potential well 250 pm wide, is in its ground state. How much energy must it absorb if it is to jump up to the state with ?

𝐸𝑛=(h2

8𝑚𝐿2)𝑛2

Δ 𝐸=𝐸4−𝐸1

52. What is the ground-state energy of (a) an electron and (b) a proton if each is trapped in a one-dimen-sional infinite potential well that is 200 pm wide?

𝐸𝑛=(h2

8𝑚𝐿2)𝑛2