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e1f10 - Version PREVIEW Exam 1 JOHNSON(53140 This print-out...

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Version PREVIEW – Exam 1 – JOHNSON – (53140) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Brodbelt 2008 photon1 001 10.0 points Compared to a 280 nm photon, a 320 nm photon has: 1. longer wavelength, lower frequency, lower energy correct 2. longer wavelength, lower frequency, higher energy 3. shorter wavelength, lower frequency, higher energy 4. shorter wavelength, lower frequency, lower energy 5. longer wavelength, higher frequency, higher energy 6. shorter wavelength, higher frequency, lower energy Explanation: Work function 01 002 10.0 points We conduct an experiment by shining 500 nm light on potassium metal. This causes electrons to be emitted from the surface via the photoelectric effect. Now we change our source light to 450 nm at the same intensity level. Which of the following is the result from the 450 nm light source compared to the 500 nm source? 1. No electrons would be emitted from the surface. 2. The same number of electrons would be emitted, but they would have a higher velocity correct 3. The same number of electrons would be emitted, but they would have a lower velocity 4. Fewer electrons would be emitted from the surface. 5. More electrons would be emitted from the surface. Explanation: 500 nm light has more energy than than the work function of potassium due to the fact that electrons were emitted. Therefore 450 nm light, which is higher in energy than the 500 nm light, will also emit electrons. The number of electrons emitted must be the same because the intensities (photons/s) are the same. However, the higher energy photons from the 450 nm light would yield electrons with a higher kinetic energy and therefore a higher velocity. 1 2 mv 2 = - Φ ChemPrin3e T01 29 003 10.0 points What is the shortest-wavelength line in the emission spectrum of the hydrogen atom? 1. 91.2 nm correct 2. 122 nm 3. 182 nm 4. 100 nm 5. 1.00 nm Explanation: The Rydberg formula gives the frequencies of the emission lines in the hydrogen atom: ν = R parenleftbigg 1 n 2 1 - 1 n 2 2 parenrightbigg . The wavelengths would be given by λ = c ν = c R parenleftbigg 1 n 2 1 - 1 n 2 2 parenrightbigg
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Version PREVIEW – Exam 1 – JOHNSON – (53140) 2 The smallest value of λ would have the largest value of ν and the largest value of 1 n 2 1 - 1 n 2 2 . This will happen when n 1 = 1 and n 2 = , giving λ = 3 × 10 8 m / s (3 . 29 × 10 15 Hz) parenleftbigg 1 - 1 2 parenrightbigg = 3 × 10 8 m / s (3 . 29 × 10 15 Hz) (1) = 9 . 11854 × 10 - 8 m because 1 2 = 1 = 0 . JB Particle Box 07 004 10.0 points An electron is in the first quantum level in a box of length L. If the box is stretched so that the length doubles, what happens to the resulting energy of the electron? How would this influence the wavelength of a pho- ton needed to excite the electron to the second energy level?
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