view-notes-pcs110-012 - CHAPTER 39 INSIDE ATOMS AND NUCLEI...

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CHAPTER 39 INSIDE ATOMS AND NUCLEI ActivPhysics can help with these problems: Activities 17.3–17.6 Section 39-1: Toward the Quantum Theory Useful numerical values of Planck’s constant, in SI and atomic units, are: h = × = × = - - 6 626 10 4136 10 34 15 . . J s eV s 1240 eV nm c / , and = = × = × - h = 2 1055 10 6 582 34 p . . J s 10 197 3 16 - = eV s MeV fm c . / . Problem 1. Find the energy in electron-volts of (a) a 1.0-MHz radio photon, (b) a 5 0 10 14 . × -Hz optical photon, and (c) a 3 0 10 18 . × -Hz X-ray photon. Solution The energy of a photon (Equation 39-1) is E hf g = , where h is Planck’s constant. For the frequencies given, the photon energies in atomic units are (a) 4.14 neV (b) 2.07 eV, and (c) 12.4 keV. Problem 2. What is the wavelength of a 6.5-eV photon? In what spectral region is this? Solution l g = = = = c f hc E = = = ( ) ( . ) . 1240 6 5 191 eV nm eV nm (We used Equation 39-1 to relate the frequency to the photon energy and atomic units for hc ). This photon is in the ultraviolet region of the electromagnetic spectrum. Problem 3. A microwave oven uses electromagnetic radiation at 2.4 GHz. (a) What is the energy of each microwave photon? (b) At what rate does a 625-W oven produce photons? Solution (a) E hf g m = = × = × - - ( . )( .4 ) . . ), 10 2 9 93 159 10 15 24 eV s GHz eV (or J as in Problem 1. (b) P= E g = ( / ) ( . / ) . 625 10 3 94 10 24 26 J s J photon = × = × - photons/s is the power output in photons. Problem 4. A red laser at 650 nm and a blue laser at 450 nm emit photons at the same rate. How do their total power outputs compare? Solution The ratio of the photon energies is E E f f blue red blue red red blue = = = = = = = = l l 650 450 1.44. Since photons are emitted at the same rate, this is also the ratio of the power outputs. Problem 5. Find the rate of photon production by (a) a radio antenna broadcasting 1.0 kW at 89.5 MHz, (b) a laser producing 1.0 mW of 633-nm light, and (c) and X-ray machine producing 0.10-nm X rays with a total power of 2.5 kW. Solution The rate of photon emission is the power output (into photons) divided by the photon energy, P P P = = = E h f hc g l = = . For
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CHAPTER 39 915 the devices specified, rates are (a) 1 6 626 10 89 5 169 10 34 28 1 kW J s MHz s = ( . . ) . , × × = × - - (b) ( ) 1 633 mW nm × ÷ ( . / ) . , 6 626 10 3 10 318 10 34 8 15 1 × × × = × - - J s m s s and (c) 1. s 26 10 18 1 × - . Problem 6. Calculate the wavelengths of the first three lines in the Lyman series for hydrogen. Solution From Equation 39-3, l = - = - = - - [ ( )] ( ) R n R n n H H 1 1 1 122 2 1 1 2 2 = = nm, 103 nm, and 97.2 nm for the first three n -values in the Lyman series, n = 2 3 , , and 4. (Note: R H = - 0 01097 1 . ( ) nm or R H - = 1 912 . nm, which is the Lyman series limit.) Problem 7. Which spectral line of the hydrogen Paschen series ( ) n 2 3 = has wavelength 1282 nm? Solution The wavelengths in the Paschen series for hydrogen are given by Equation 39-3, with n n 2 1 3 4 5 6 = = and . , , , . . . Thus, l = - ( ) ( ). 1 9 9 1 2 1 2 = = R n n H If one substitutes l = 1282 nm and R H = - 0 01097 1 . ( ) , nm one finds 9 0 01097 1 2 n = × . 1282 9 1 2 ( ), n - or n 1 9 141 9 5 = × - = ( . ) ( . ) , = so this is the second line in the series.
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This note was uploaded on 10/19/2010 for the course PHY 78647 taught by Professor Miller during the Spring '09 term at Albany College of Pharmacy and Health Sciences.

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view-notes-pcs110-012 - CHAPTER 39 INSIDE ATOMS AND NUCLEI...

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