Final Exam answers relevant to Summer 2014 Exam 2 I. Relativity, Quantum, Atomic (choose 3 of 4)

Size: px

Start display at page:

Download "Final Exam answers relevant to Summer 2014 Exam 2 I. Relativity, Quantum, Atomic (choose 3 of 4)"

Claire Patterson
5 years ago
Views:

1 PH 02 / LeClai Summe II 200 Final Exam answes elevant to Summe 204 Exam 2 I. Relativity, Quantum, Atomic (choose 3 of 4). In a scatteing expeiment to eveal the atomic-scale stuctue of a mateial, electons ae acceleated though a potential diffeence of V = 500 kv. What is the smallest featue one could hope to see befoe quantum uncetainty spoils the esolution? Solution: The potential diffeence gets you potential enegy, potential enegy change equals kinetic enegy change. Kinetic enegy gets you momentum, momentum gets you wavelength. KE = P E = e V = p2 2m = p = 2me V () λ = h p = h m (2) 2me V 2. What ae the possible wavelengths of photons that could be emitted fom a system with an enegy level diagam like that in the figue below? 9.36 E (ev) Solution: The possible photon enegies ae the possible enegy diffeences, = 5.2 ev, =3.2 ev, =8.32 ev. 3. A muon is fomed fom a cosmic ay showe in the uppe atmosphee at an altitude of km. In its own efeence fame, the muon has a lifetime of s. (a) If the muon moves at 0.98c, what is its mean lifetime as measued by an obseve on eath? (b) Will it each the suface of the eath?

2 (c) If you ignoed elativity, would it each the suface of the eath? 4. The kinetic enegy of an electon is inceased by a facto of two. Neglecting elativistic effects, what facto does its wavelength change? Solution: Use the kinetic enegy-momentum elationship: Now use the wavelength-momentum elationship: K = p2 2m = p = 2mK (3) λ = h p = If kinetic enegy doubles, wavelength must decease by a facto 2. h 2mK (4)

3 II. Electic foces, fields, and enegy (choose 3 of 4) 5. The two figues below show small sections of two diffeent possible sufaces of a NaCl suface. In the left aangement, the NaCl(00) suface, chages of +e and e ae aanged on a squae lattice as shown. In the ight aangement, the NaCl(0) suface, the same chages ae aanged in a ectangula lattice. What is the electical potential enegy of each aangement (symbolic answe only)? Which is moe stable? a + - a a a 2 6. A conducting spheical shell of adius suounds a point chage +q. Outside the shell, a thee is a point chage q. At point P, a distance below the q chage and 2 to the ight of the +q chage, what is the magnitude of the total electic field? -q +q P conducto Solution: Gauss law tells us that the point chage +q and the spheical shell can be teated as single point chage +q. The conducto doesn t matte so long as you ae outside of it. That means the field at P is just that of two point chages added togethe: one field E + due to a chage +q at a distance 2, the othe field E due to a chage q at a distance R. E + = k eq (2) 2 = k eq 4 2 (5) E = k eq 2 (6)

4 Howeve, since they act at ight angles to one anothe, we have to add them as vectos. This is simple enough due to the fact that they ae pependicula: E tot = E+ 2 + E 2 = k eq = 7 4 k e q 2 (7) 7. A household lamp has a cod with wie of diamete mm and uses a total of 3 m of wie to connect to a souce of voltage. If the lamp caies a steady cuent of 0.5 A, and the coppe has a chage density of n= electons/m 3, how long would it take fo one electon to tavel the full length of the cod? Solution: The length of time it will take is dictated by the dift velocity of the electons v d. We know this is elated to the cuent: I = nqv d A o v d = I nqa We ae given n, and fom the wie diamete d we can find the coss-sectional aea A (8) A = π 2 = 4 πd2 (9) This is now enough to detemine the velocity. At a velocity v d the electons will cove a distance l =v d t in a time t. If l is the length of the wie, l = v d t = t = nqal I I nqa t (0) = nqπd2 l 4I sec 8 h () 8. How many electons should be emoved fom an initially unchaged spheical conducto of adius m to poduce a potential of 7.50 kv at the suface?

5 III. Cicuits (choose 2 of 3) 9. The cicuit below is a cude electic eye I built a few nights ago. Its pimay components ae a photoesisto (R 2 ), and a light-emitting diode (LED). The photoesisto has the popety that in the dak, its esistance is about R 2,dak = 50 kω, while in bight light its esistance is damatically lowe, about R 2,light = 0.3 kω. The (ed) LED in the cicuit will light up when its voltage dop exceeds.8 V. You may assume that the light fom the LED does not each the photoesisto to change its esistance. Take R =3 kω and R 3 = 370 Ω. The LED has a esistance of 75 Ω when lit, and MΩ when not. When the photoesisto is exposed to bight light (i.e., in its low esistance state), should the LED be lit o not? Explain you easoning. Hint: the voltage acoss the LED is mostly detemined by the potential dop acoss R, which is detemined by the net cuent flowing out of the battey. R R V R2 LED Solution: Not eally elevant to the summe 204 exam 2, but hee s a ough qualitative explanation. When the photoesisto is lit, its esistance is vey low, and it takes a lot of cuent fom the battey (compaed to the much highe R 3 and diode esistances in seies). If the photoesisto esistance becomes too low, the battey can t supply enough powe to light the LED. When the photoesisto is dak, it has a high esistance, and takes little cuent fom the battey, leaving it plenty of capacity to light the LED. 0. Refe to the pevious question and its figue. Ignoe the LED (petend it is just a wie), and let R =300 Ω and R 3 =200 Ω. The photoesisto behaves the same as in the pevious question. (a) What is the voltage on R 3 when the photoesisto is dak? (b) What is the voltage on R 3 when the photoesisto is illuminated? Hint: note that R 2 and R 3 ae in paallel.. The cicuit on the next page is pat of an instument in my laboatoy, a Keithley Instuments model 428 cuent amplifie. The section of the instument in the diagam below is a filte with a selectable fequency cutoff. The fequency ange is selected by opening o closing vaious switches shown in the diagam. The uppe thee switches (U, U2, and U3) selectively bypass esistos R34, R50, and R38 to vay the seies esistance. Switches U06, U5 allow one to

6 connect vaious capacitances to gound befoe the signal output. (The two switches labeled U 06 open and close togethe.) (a) What sot of filte is this? Justify you answe in a sentence o two. (b) What is the cutoff fequency when switches U5 and U ae closed, but all othe switches ae open? N.B. - the inveted tiangles in this diagam ae gound connections. Recall p = 0 2, µ = 0 6, and k =0 3. Solution: This is a low-pass filte. The capacitos pesent a low eactance to high fequencies, and take them to gound. Low fequencies see the capacitos as a high eactance, and ty to avoid

7 U U2 U3 Figue : A potion of the input filte in a Keithley 428 cuent peamplifie. Fom the K428 instument manual. them, so they make it to the output. If U5 is closed, that connects capacitos C3 and C32 to gound, but no othes. In paallel, they have an equivalent capacitance of 74.8 nf (watch the metic pefixes... ). If U is closed, that shots out esisto R34, leaving esistos R50, R38, and R32 in seies, which have an equivalent esistance of 4.07 kω. Thus, with this combination of switches closed, the low-pass filte is one with R = 4.07 kω and C = 74.8 nf. The cutoff fequency is when the two have equal eactances: X C = X R (2) 2πfC = R (3) f = 52 Hz 2πRC (4) (5) IV. Magnetism, induction (choose 2 of 3) 2. A wie caies a cuent I = 5 A to the east. A distance d diectly below it, an electon tavels west at m/s. (a) What is the foce on the electon in magnitude, and is it attacted o epelled fom the wie? (b) If the electon is to tavel in a staight line paallel to the wie, what is d? 3. A lage solenoid of length.0 m, a adius of 0.0 m, and N = 000 tuns caies a cuent of I = 40 A and fully encloses a smalle solenoid of length 0.5 m, adius of m, and N 2 = 5000 tuns (see the cude figue on the following page). The lage solenoid ceates a unifom magnetic field ove the aea of the smalle solenoid. Duing a powe outage, the cuent in the oute coil suddenly deceases to zeo ove a time t=0. s. What voltage is induced between the end points of the inne solenoid?

8 4. Thee wies cay a cuent I, as shown in the figue below (the wies coss but do not touch). The vetical wie cosses the paallel hoizontal wies at a 90 angle, and the two hoizontal wies ae a distance d apat. What is the magnetic field at point P, a distance d fom the vetical wie and halfway between the hoizontal wies? Solution: Fom the second ight hand ule, we can figue out that the field fom all thee wies is into the page at point P. Since they all have the same diection, we just have to add the thee fields togethe. We have two wies at a distance d/2 and one at a distance d, all caying a cuent I, so the total field is B tot = B + B 2 + B 3 = µ oi 2π(d/2) + µ oi 2π(d/2) + µ oi 2πd = µ oi 2πd ( ) = 5µ oi 2πd (6)

9 V Name & CWID V I d I d x P I

10 V. Optics, EM waves (choose 2 of 3) 5. A small object is placed at a distance of 0 cm fom a conveging lens, and an image is found to fom on the opposite side of the lens a distance of 8 cm fom the lens. (a) What is the focal length of the lens? (b) What is the magnification facto? Solution: We know the image and object distances, focal length is no poblem: p + q = f = = f = 4.44 cm (7) Magnification is just M = q/p= 8/0= Two paallel mios of height h = m ae a distance d = 0. m apat. A light ay entes though the bottom at an angle of 30. How many times will the incident beam be eflected by each of the paallel mios? d h 30 o Solution: Just geomety. Each time the light bounces once, it has moved upwad by an amount δ =d tan 30. If the light bounces off n times, it has moved upwad by nδ, which needs to be geate to o equal than h. Thus, nδ = h (8) n = h δ = 7 bounces won t quite do it, 8 ae equied. h = 7.3 (9) d tan The intensity of moonlight when it eaches the Eath s suface is appoximately 0.02 W/m 2 (fo a full moon). What ae the amplitudes (maximum intensity) of the coesponding electic and magnetic fields? Solution: We can wite intensity in tems of eithe the E o B field maxima:

11 I = E m 2µ o c = cb2 m 2µ o (20) With the known intensity, fist find one and then the othe (o, find one and note E = cb fo an EM wave). E m 3.88 V/m, B m =E m /c 3 nt.

12 Constants: g 9.8 m/s N A = things/mol k e = = N m 2 C 2 4πɛ o µ o 4π 0 7 T m/a ɛ o = 4πk e = C 2 /N m 2 e = C h = J s = ev s = c = h 2π = m/s µ0 ɛ 0 m e = kg = MeV/c 2 m p + = kg = MeV/c 2 m n 0 = kg = MeV/c 2 u = MeV/c 2 hc = ev nm EM Waves: c = λf = E B [ photons I = time I = Electic Potential: ] [ ] [ enegy photon Aea enegy time aea = EmaxBmax powe (P) = = E2 max 2µ 0 aea 2µ 0 c ] V = V B V A = PE P E = q V = q E x cos θ = qe x x constant E field q V point chage = k e q q 2 P E pai of point chages = k e 2 P E system = sum ove unique pais of chages = W = PE = q(v B V A ) q pais ij k eq i q j ij Quadatic fomula: 0 = ax 2 + bx 2 + c = x = b ± b 2 4ac Basic Equations: F net = m a Newton s Second Law 2a F cent = mv2 ˆ Centipetal Magnetism F B = q v B sin θ vb F B = BIl sin θ wie F 2 l µ 0 I B = 2π wie B = µ 0 I loop 2 N B = µ 0 L I ẑ µ 0nI ẑ solenoid = µ 0 I I 2 2πd Cuent & Resistance: I = Q t = nqav d 2 wies, foce pe length v d = eτ E τ = scatteing time m m ϱ = ne 2 τ V = ϱl A I = RI R = V = ϱl I A P = E t = I V = I2 R = [ V ] 2 powe R Optics/Photons: E = hf = hc λ speed of light in vacuum n = speed of light in a medium = c v λ = v = c/n = n 2 efaction λ 2 v 2 c/n 2 n n sin θ = n 2 sin θ 2 Snell s efaction λf = c M = h h = q p f n p + n 2 q = p + q = 2 R = n 2 n R q = n 2 p n f = spheical mio & lens spheical mio & lens spheical efacting flat efacting ( n2 n n ) [ R R 2 ] Electic Foce & Field F e,2 = q E 2 = keq q ˆ 2 Capacitos: E = k e q 2 Φ E = E A cos θ EA = Q inside ɛ 0 lensmake s P E = W = q E x cos θ = qe x x constant E field Q capacito = C V C paallel plate = ɛ 0A E capacito = Q2 Q V = 2 2C C eq, pa = C + C 2 C eq, seies = d + C C 2 C with dielectic = κc without 2

13 Resistos: I V souce = V ated R + R V V souce = V ated + R I I souce = I ated + R R eq, seies = R + R 2 R eq, pa = + R R 2 Right-hand ule #. Point the finges of you ight hand along the diection of v. 2. Point you thumb in the diection of B. 3. The magnetic foce on a + chage points out fom the back of you hand. Right-hand ule #2: Point you thumb along the diection of the cuent (magnetic field). You finges natually cul aound the diection the magnetic field (cuent) ciculates. Vectos: Induction: F = Fx 2 + F y 2 [ ] magnitude θ = tan Fy diection F x Φ B = B A = BA cos θ BA V = N Φ B t L = N Φ B I = NΦ B I V = v B l = E l motional voltage ac Cicuits τ = L/R RL cicuit Relativity τ = RC RC cicuit X C = 2πfC esistance of a capacito fo ac X L = 2πfL esistance of an inducto fo ac ω cutoff = γ = τ = 2πf cutoff v2 c 2 t moving = γ t stationay = γ t p L moving = L stationay γ t = t t 2 = γ ( t v x c 2 ) v obj = v + v obj + vv obj c 2 KE = (γ )mc 2 E est = mc 2 p = γmv E 2 = p 2 c 2 + m 2 c 4 v obj = v obj v vv obj c 2 Unit Symbol equivalent to newton N kg m/s 2 joule J kg m 2 /s 2 = N m watt W J/s=m 2 kg/s3 coulomb C A s amp A C/s volt V W/A = m2 kg/ s3 A faad F C/V = A 2 s4 /m 2 kg ohm Ω V/A = m2 kg/s3 A2 tesla T Wb/m 2 = kg/s 2 A electon volt ev J - T m/a N/A 2 - T m 2 V s - N/C V/m Powe Pefix Abbeviation 0 2 pico p 0 9 nano n 0 6 mico µ 0 3 milli m 0 2 centi c 0 3 kilo k 0 6 mega M 0 9 giga G 0 2 tea T 3

Phys 1215, First Test. September 20, minutes Name:

Phys 1215, First Test. September 20, minutes Name: Phys 115, Fist Test. Septembe 0, 011 50 minutes Name: Show all wok fo maximum cedit. Each poblem is woth 10 points. k =.0 x 10 N m / C ε 0 = 8.85 x 10-1 C / N m e = 1.60 x 10-1 C ρ = 1.68 x 10-8 Ω m fo