Experiment 1 - Lenses and Geometric Optics 1. The refractive index of water is nw= 1.330. Diamond has a greater refractive index of (n=2.5) which is why it sparkles. Its light can only show over a small range of angles. 2. 0.018m, which is smaller. Experiment 2 - Wave Nature of Light 1. Calculate the slit width of a single slit by analyzing the diffraction pattern. List possible sources of error in your answer. 2. Wavelength calculation for double slit and the diffraction grating; A. Use the interference pattern from the double slit to calculate the wavelength of the laser. Compare this with the given value. B. Repeat the wavelength calculation using the diffraction grating. Discuss the differences in the pattern of the screen when using the grating and the double slit. Which one makes a more accurate measurement? Experiment 3 - DC Circuits and Ohm's Law 1. Think about a string of holiday lights. When one bulb burns out, the other bulbs continue to shine. What does that mean for the nature of the wiring of the string? If a bulb burns out while the others stay on, the string of lights is parallel wired. 2. Consider two resistors of equal resistance. You can combine them in a circuit either in series or in parallel. Which combination would have the lower effective resistance, series or parallel? Which would carry more current: series or parallel? Parallel: lower effective resistance. Bulbs in series carry more current. Experiment 4 - Magnetism and Induction 1. B=.6T A=.02m2 n=200 emf=2.4Vmf e = − nΔtΔBA B=.6T A=.02m2 emf = − n n=200 emf=2.4V Δt ΔBA 2. 2. No. There is no flux in the field. there is no change to record. Experiment 5 - Atomic Spectra 1. Calculate the wavelength of light emitted when an electron in a hydrogen atom makes a transition from n=7 to n=3.-2.179E-18 • (⅓^2) - (1/7^2) -2.179E-18 • 0.1111 - 0.0204 -0.1974 / 6.626 E-34 2.998E8 / -2.97943E32 = -1.00623E-24 Infrared, not visible. 2. Use the same equation to calculate the wavelength in nm of light emitted when the electron in a hydrogen atom undergoes a transition from n = 3 to n = 2. -2.179E-18 • (½^2) - (⅓^2) -2.179E-18 • 4 - 0.1111 -0.000000000000000008474131 / 6.626 E-34 2.998E8 / -1.279E16 = -2.34 e-8 Experiment 6 - Radioactivity 1. A radioactive element has an atomic mass number (A) of 241 and an atomic number (Z) of 95. How does the atomic mass number and the atomic number change when the element disintegrates into: A | Z a) An particle 237 | 93α b) A * particle 241 | 96β c) A particle 241 | 96β 2. Polonium-210 is an alpha (2_He^4) emitter. It decays in the following way: 84_Po^210 > Pb+ 2_He^4 Lead (Pb) is the daughter nuclei. Using the decay above calculate Pb mass number (A). What is Pb proton number (Z)? A | Z 207.2
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