Download Physics Exam 2: Birth of Quantum Mechanics and more Exams Advanced Physics in PDF only on Docsity! NAME: Febraury 19, 2002 PHGN310: Modern Physics Exam 2: Birth of Quantum Mechanics Instructions: 50 minute exam, closed book, one page of notes, calculators permitted. Show ALL work. Use the back of the exam if additional space is needed. I. (20) A ball bearing .25 cm in radius is heated to 2500 ◦K. Assuming it can be treated as a blackbody, approximately how many photons per second are emitted between the wavelenths of λ = 695 nm and λ = 705 nm? (HINT: Since the wavelength band is narrow, you don’t have to do any integrals.) SOLUTION: The Planck radiation formula gives the radiated power per area per wavelength. I(λ, T ) = 2πhc2 λ5 1 eE(λ)/kT−1 . Normally, to get the power per area over many different wavelengths, you would have to integrate with respect to wavelength, but for this problem we just need an approximate answer; so, since the wavelength range is so narrow, all you only multiply the intensity per wavelength at the average wavelength (I(λ = 700 nm, T = 2500 ◦K)) times the wavelength band (10 nm). Some students incorrectly used the Stefan-Boltzmann law which is the power per area for ALL wavelengths. To get the total power multiply by the area of the ball bearing (4πR2, not πR2)). The approximate number of photons emitted per second is the energy of one photon at the average wavelength (E = hc/(700)nm) divided into the total power. (For numerical convenience you can use the fact that room temperature (300 ◦K) is approximately 1/40 eV.) II.(20) An electron microscope “shines” electrons of kinetic energy 250 eV on a sample. The electron microscope is attempting to image a cubic crystal with a lattice constant of 0.128 nm. The crystal is oriented so the crystal planes are parallel to the surface of the sample. At what grazing angles between 0 and 90 degrees will there be interference maxima? SOLUTION: Constructive interference occurs when 2d sin(θ) = nλ so θ = sin−1(nλ/(2d)). The deBroglie wavelength is given by λ = h/p = hc/(pc) where the momentum is obtained from the kinetic energy by K = p2/(2m) = (pc)2/(2mc2) or pc = √ 2mc2K. Plugging in the numbers gives three possible angles between 0 and 90 degrees. After that the argument of the arcsin is greater than one which is unphysical. IV.(20) In advanced lab a group of students scattered a well collimated beam of 5 MeV α-particles from a zirconium foil 2 µm thick. In a 15 minute counting period they detected 4,000 counts when their detector was at an angle of 60 degrees. (ZZr = 40, atomic mass of Zr is 91.2, density of Zr: ρ = 6.51 g/cm3, NA = 6.022× 1023) (a) Estimate the number of counts they should get when their detector is located at an angle of 45 degrees. SOLUTION: The number of alphas scattered at some angle, N(θ), is inversely proportional to the fourth power of the sine of half the scattering angle: N(θ) ∝ sin−4(θ/2). Thus: N(θ1) = N(θ2)[ sin(θ2/2)/ sin(θ1/2) ]4. (b) Estimate the number of counts they should get in their detector at 60 degrees when their target is replaced by a 1 µm thick silver foil? (ZAg = 47, atomic mass of Ag is 107.87, density of Ag: ρ = 10.5 g/cm3) SOLUTION: The number of alphas scattered by a target is proportional to the charge (atomic number,Z), the number density, and the thickness of the target. Thus: N1 = (Z1/Z2)2(n1/n2)(t1/t2)N2. The number density is given by n = ρNA/Matomic. IV. (20) A 500 MeV photon scatters 135 degrees off a proton (NOT an electron) initially at rest. With how much energy does the proton recoil? SOLUTION: The initial wavelength is given by the energy: λ = hc/Eγ . The compton scattering formula for the shift in wavelength is: ∆λ = λ′−λ = λc(1−cos(θ)) where λc = hc/(mpc2) is the compton wavelength of the particle (proton in this case) from which the photon scatters. From the final wavelength of the photon calcluate the final energy of the photon: E′ = hc/λ′. The energy lost by the photon is the (kinetic) energy gained by the proton: Kp = Eγ − E′γ . 1
