Mid Term Exam 1. Feb 13, 2009

Transcription

1 Name: ID: Mid Term Exam 1 Phys 48 Feb 13, 009 Print your name and ID number clearly above. To receive full credit you must show all your work. If you only provide your final answer (in the boxes) and do not show your work you will only receive partial credit, even for a correct final answer. Problem Score 1 /0 /0 3 /0 4 /30 Total /90 Useful constants: Planck constant h = 6.66 x J s Reduced Planck constant = h =1.054 x J s hc = 140 ev nm h c = ev nm c = 3 x 10 8 m/s 1 ev = 1.6 x J m e = electron mass = 9.11 x kg = MeV

2 Problem 1: short problems. 1. A bat can detect very small objects, such as insects whose length is approximately equal to one wavelength of the sound the bat makes. If the bat emits chirps at a frequency of 55.0 khz, and the speed of sound in air is 340 m/s, what is the smallest insect the bat can detect? d = " = v f = #10 3 = 6.18mm d. A particle executes simple harmonic motion with an amplitude of 4.00 cm. At what position does its speed equal half its maximum speed? v max = "A v = "A E = E k + U # 1 ka = 1 mv + 1 kx # " A = v + " x = " A x = 3 4 A # x = ± 3 A = 3.46cm 4 + " x # x 3. A seismic wave far from the epicenter of an earthquake can be modeled as a wave on a string transporting energy (we neglect any absorption). The wave moves from granite into mud fill with same density but lower bulk modulus. The speed of the wave drops by a factor of 5.0 with negligible reflection of the wave. Will the amplitude of the ground shaking increase or decrease? By what factor?

3 The rate of energy transfer is: P = const va => v granite A granite = v mudfill A mudfill => A mudfill = A granite v granite v mudfill = 5A granite Increase A mud fill = x A granite 4. Standing wave vibrations are set up in a crystal goblet with 4 nodes and 4 antinodes around the 0.0 cm circumference of its rim. If transverse waves move around the goblet at m/s what would be the high harmonic frequency an opera singer has to produce to shatter the goblet with a resonant vibration? d NN = " = 0 4 = 5.0cm # " =10.0cm f = v " = 450 = 4.5kHz % 10 $10 f Problem : multiple choice. Circle your answers. 1. The graph shows the kinetic energy of photoelectrons ejected from a certain metal as a function of the frequency of the incident light on the metal. What is the work function of the metal? (a) 50 ev (b) ev (c) 5.0 x 10-0 J (d) 0 ev (e) 33 x ev (b) The work function is the point where the line K max = hf " # intercepts the y axis, that is K max = -Φ. From the graph the line has to go through

4 (x,y) = (5.0x10 14 Hz, 0) 0 = hf- Φ 0 = 6.66 "10 #34 " 5.0 "10 14 # $ % $ = "10 #0 J =.07eV. The circular pupil of the human eye has a diameter of 4.0 mm. The minimum light intensity that the eye is sensitive to is 1.0 x W/m. Approximately how many photons per second of wavelength 550 nm does this correspond to? A) 350 s -1 B) s -1 C) 3.5 x 10 1 s -1 D) 3500 s -1 E) 3.5 x 10 - s -1 # photons second Power " Area Power " Area = = = Energy / photon hf Power " Area " # hc = = 1"10$10 " % " (4.0 "10 $3 /) " eV & nm "1.6 "10 $19 = s $1 ' 3500s $1 D)

5 3. Find the de Broglie wavelength of a beam of electrons with kinetic energy 100 ev. A) nm B) nm C) nm D) 0.1 nm E) nm Electrons with E k = 100 ev are not relativistic since E k << mc. p = D) mc E k c " # = h p = 140eV $ nm % 0.511%10 6 %100 = 0.1nm 4. The beam of electrons in #3 is used in a Davisson-Germer experiment. The beam travels toward a crystal of inter-planar spacing d = 3.5 Å. Compute the angle θ where there is a strong reflected peak that is equal to the incidence angle (see figure). d (a) 50 o (b) 80 o (c) 45 o (d) 60 o (e) 65 o According to the figure the path difference between the two beams is: dcos"

6 Constructive interference: % $ ( d cos" = m# $for m=1 " = cos #1 ' * = o + 80 o. & d ) (b) Problem 3: single slit diffraction. A diffraction pattern is formed on a screen 10 cm away from a mm wide slit. Monochromatic nm light is used. a) What is the width of the central maximum? b) Calculate the fractional intensity I/I max at a point on the screen 4.10 mm from the center of the principal maximum. a) Width tan! min = "y L # sin! = $ L$ min % "y = a a = 10cm & & 10'7 cm 0.04cm b) The distance Δy corresponds to the angle: sin" # tan" = $y L = 0.38cm

7 The total phase difference between waves from the top and bottom of the slit is: " = # $ asin% and the intensity at a point on the screen divided by the maximum intensity corresponding to θ=0 is # sin(" /)& I /I max = % ( $ " / ' # sin()asin* /+)& = % ( $ ()asin* /+) ' # sin()a,y /(L+))& = % ( $ ()a,y /(L+)) ' = I/I max Problem 4: wave function and probability. The wave function of an electron in a box is $ "(x) = Asin 3#x ' & ) % L ( for 0 x L and zero elsewhere, where A is a normalization constant and L = 10 nm. a) If we perform an experiment to measure the location of the electron, what is the probability that we find it between x=l/3 and x = L/? b) If the energy of the electron is measured, what is its value? c) Estimate the energy of the fundamental state by means of the uncertainty principle. The probability to find the particle between 0 and L is 1, hence we determine A: L L L % 3$x ( L 1 1= # " * " dx = A # sin ' * dx = A # ( 1+ cos( y) ) dy = & L ) 3$ 0 = A L 6$ 3$, A = L 0 0

8 Probability L / P = # " * " dx = L L / 3 % 3$x ( sin ' * dx = 1 L / % % # 1+ cos 6$x (( # ' ' ** dx = & L ) L & & L )) L / L / 3 = 1, L L + L 3 + L (sin 3$ -. 6$ ( ) + sin$) / 0 1 = 1 6 L / 3 b) The particle is in the n=3 state, so E = E 3 = (3)! h = 9! h ml ml. Numerically E 3 = 3.38e- ev Energy c) Since the electron is confined in the box: "x = L # "p x $ h L ("p x ) = ( p x % p x ) = p ( x % p x p x + p x ) ave Since the probability that the electron goes in one direction or the opposite is the same, the average momentum along the dimension of the box is zero: ("p x ) = p x # E = p x m $ 1 % h( ' * # E m & L) min = 1 % h( ' * m & L) = ,3 J = ,4 ev E min