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hw12solutions

hw12solutions - Physics 214 Homework 12 Solutions Spring...

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1 Physics 214 Homework 12 Solutions Spring 2008 Problem 1 a) The wavelength of the gamma rays is m 10 1.24 eV) J 10 (1.60 eV) 10 (1.00 ) s m 10 (3.00 s) J 10 (6.63 λ 12 19 6 8 34 - - - × = × × × × = = i i E hc and the wavelength of the final visible photons is m. 10 00 . 5 7 - × So, the increase in wavelength per interaction (assuming about 26 10 interactions) is . m 10 00 . 5 10 m 10 1.24 m 10 00 . 5 33 26 12 7 - - - × = × - × b) ). small for ( 2 2 1 1 ) cos 1 ( λ 2 2 φ φ φ φ mc h mc h mc h = + - - = Δ . ) 10 (3.68 radians 10 6.42 s) J 10 (6.63 m) 10 (5.00 ) s m 10 (3.00 Kg) 10 2(9.11 λ 2 So 9 11 34 33 8 31 ° × = × = × × × × = Δ - - - - - h mc φ c) × = × × = sec collisions 10 3.17 sec 10 3.15 collisions 10 sec 10 3.15 ears 10 12 13 26 13 6 y . collision sec 10 15 . 3 13 - × How far does light travel in this time? = × × - ) collision sec 10 (3.15 ) s m 10 00 . 3 ( 13 8 . 0.1mm mm 0.0945 m 10 45 . 9 5 = × - Problem 2 a) f microwave = 2.45 GHz (microwave frequency EM waves—what a surprise!) f cellphone = 850 MHz (radio/microwave frequency EM waves) f powerline = 60 Hz (VERY long radio frequency EM waves) Applying the formula hf E = and converting from Joules to eV for each object, then finding the ratio of eV 1 = bond E to the energy of each object gives: eV 10 02 . 1 eV) J 10 (1.60 ) s 10 s)(2.45 J 10 (6.63 5 19 -1 9 34 - - - × = × × × = = microwave microwave hf E 4 5 - 10 85 . 9 eV 10 1.02 eV 1 × = × = microwave bond E E Thus the energy needed to break chemical bonds is about 100,000 times greater than the energy of the photons released by the microwave. eV 10 52 . 3 eV) J 10 (1.60 ) s 10 s)(850 J 10 (6.63 6 19 -1 6 34 - - - × = × × × = = cellphone cellphone hf E

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2 5 6 - 10 84 . 2 eV 10 3.52 eV 1 × = × = cellphone bond E E Thus the energy needed to break chemical bonds is about 300,000 times greater than the energy of the photons released by your average cell phone. eV 10 49 . 2 eV) J 10 (1.60 ) s s)(60 J 10 (6.63 13 19 -1 34 - - - × = × × = = powerline powerline hf E 12 13 - 10 02 . 4 eV 10 2.49 eV 1 × = × = powerline bond E E Thus the energy needed to break chemical bonds is about 4 trillion times greater than the energy of the photons released by power lines. b) Since we are trying to be very thorough, we also consider the excitation of molecules so that their chemical bonds stretch and twist, but do not break; this type of excitation is called vibrational excitation. The energy l vibrationa E can be related to the thermal temperature scale T k B to determine at what temperature thermal energy can cause vibrational energy excitations. Vibrational levels for most molecules begin to be excited by thermal energies at temperatures around 1000 Kelvin. Thus, assuming about room temperature or 300 K to be around the lowest temperature at which vibrational excitation occurs, eV 026 . 0 J/eV) 10 60 . 1 ( K) 300 )( K / J 10 38 . 1 ( ~ 19 23 = × × = - - T k E B l vibrationa 3 5 - 10 54 . 2 eV 10 1.02 eV 026 . 0 × = × = microwave l vibrationa E E Thus the energy needed to excite vibrations is about 2000 times greater than the energy of the photons released by the microwave.
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