Solutions to Exam II: The Birth of Quantum Mechanics in PHGN300: Modern Physics, Exams of Advanced Physics

Download Solutions to Exam II: The Birth of Quantum Mechanics in PHGN300: Modern Physics and more Exams Advanced Physics in PDF only on Docsity! PHGN300: Modern Physics Exam II: The Birth of Quantum Mechanics October 2, 2003 NAME: KEY 1. (10) Estimate the average temperature of the surface of Mars assuming it behaves as a perfect radiator. Explain your reasoning. (DATA: the solar intensity on Mars is 410 W/m2, the radius of Mars is 3.44×106m.) Solution: When equilibrium is reached, power absorbed (in) equal power radiated (out). Power absorbed is the radiation absorbed on the cross section of mars: Pin = 410(W/m2)πR2. Power radiated is given by the Stefan-Boltzman law radiated over the entire surface area: Pout = σT 24πR2. Equating these gives: T = ( 410(W/m2)πR2 σ4πR2 )1/4 = 206◦K. (1) 2. (10) A vacuum tube has both the anode and cathode made of platinum which has a work function of 6.25 eV. When ultraviolet light hits the anode, a current is detected in an ammeter connected between the cathode and anode. This current stops when a bias voltage of 14.5 V is applied. What is the wavelength of the light? Solution: Photoelectric effect: eVstop = hc/λ − φ. This implies that hc/λ = eVstop + φ. Solving for λ gives: λ = hc/(eVstop + φ) = 59.76 nm. 3. (20) Students in the advanced laboratory conduct a scattering experiment using a 7.0 MeV alpha source and a thin gold (Z=79) foil target (.1 µm thick). Their alpha detector has an area of 0.5 cm2 and is placed 4 cm away from the target foil at an angle of 30 degrees with respect to the initial beam direction. In a one hour period the students register 8,000 counts in the detector. (a) (10) Suppose they were to replace the 7.0 MeV alpha source with a 5.0 MeV alpha source of the same activity (i.e. same number of incident alpha’s per unit time). With all other factors the same, how many counts will they get in one hour now? Solution: The cross section scales with the square of the distance of closest approach, Dc. The distance of closest approach in inversely proportional to the kinetic energy; so the cross section is inversely proportional to the square of the kinetic energy. Thus: N(5MeV ) = N(7MeV )(7MeV5MeV ) 2 = 15, 680 counts. (b)(10) Using the original 7.0 MeV alpha source, the students replace the target gold foil by a foil of an unknown metal whose thickness is engineered to have exactly the same number of target atoms in the alpha beam as the original gold foil. With the detector at 30 degrees the students register 2,051 counts in a one hour period. What is the atomic number of the unknown metal? Solution: As in part (a), the cross section scales with the square of the distance of closest approach, Dc, which is proportional to the charge of the target, Z. Thus: N(Zx) = N(ZAu)( ZxZAu ) 2. This gives Zx = ZAu( N(Zx) NAu )1/2 = 40 (Zirconium). 4. (10) A photon scatters 60-degrees from an electron in an aluminum block. The exiting photon has a wavelength of 2.3 pm (pico-meter = 10−12 m). What was the original energy of the photon? Solution: Using the Compton equation and solving for the initial wavelength: λ = λ′ − ∆λ = λ′ − λc(1 − cos θ) = 1.09× 10−3nm. Thus the initial photon energy is: Eγ = hc/λ = 1.14 MeV. 5. (50) Consider an exotic atom consisting of a proton and a kaon-minus. (The kaon has the same charge as an electron but its mass is 493.6 MeV/c2, proton mass is 938.3 MeV/c2.) (a) (10) What is the reduced mass for this system (in MeV/c2)? Solution: The reduced mass formula gives: mexotic = mKmp/(mK + m + p) = 323.5 MeV/c2. (b) (10) What is the Bohr radius of this exotic atom? Solution: The Bohr radius scales inversely with the reduced mass. The reduced mass for ordinary hydrogen (electron- proton) is very close to the electron mass (0.511 MeV/c2) because the proton is much heavier than the electron. Thus: aexotic = a0( memexotic ) = 83.7 fm. (c) (10) What is the energy of a photon emitted when this exotic atom undergoes a transition from the n = 3 state to the n = 2 state? 1