# Ch38Word - Chapter 38 p. 1 CHAPTER 38 Early Quantum Theory...

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Chapter 38 p. 1 CHAPTER 38 – Early Quantum Theory and Models of the Atom Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc / λ = (6.63 × 10 –34 J · s)(3.00 × 10 8 m/s)(10 –9 nm/m)/(1.60 × 10 –19 J/eV) ; E = (1.24 × 10 3 eV · nm)/ . 1. We find the temperature for a peak wavelength of 440 nm: T = (2.90 × 10 –3 m · K)/ P = (2.90 × 10 –3 m · K)/(440 × 10 –9 m) = 6.59 × 10 3 K . 2. ( a ) The temperature for a peak wavelength of 25.0 nm is T = (2.90 × 10 –3 m · K)/ P = (2.90 × 10 –3 m · K)/(25.0 × 10 –9 m) = 1.16 × 10 5 K . ( b ) We find the peak wavelength from P = (2.90 × 10 –3 m · K)/ T = (2.90 × 10 –3 m · K)/(2800 K) = 1.04 × 10 –6 m = 1.04 µ m . Note that this is not in the visible range. 3. Because the energy is quantized, E = nhf , the difference in energy between adjacent levels is ? E = hf = (6.63 × 10 –34 J · s)(8.1 × 10 13 Hz) = 5.4 × 10 –20 J = 0.34 eV . 4. ( a ) We find the peak wavelength from P = (2.90 × 10 –3 m · K)/ T = (2.90 × 10 –3 m · K)/(273 K) = 1.06 × 10 –5 m = 10.6 m . This wavelength is in the infrared . ( b ) We find the peak wavelength from P = (2.90 × 10 –3 m · K)/ T = (2.90 × 10 –3 m · K)/(3300 K) = 8.79 × 10 –7 m = 879 nm . This wavelength is in the near infrared . ( c ) We find the peak wavelength from P = (2.90 × 10 –3 m · K)/ T = (2.90 × 10 –3 m · K)/(4 K) = 7.25 × 10 –4 m = 0.73 mm . This wavelength is in the far infrared . 5. ( a ) The potential energy on the first step is U 1 = mgh = (58.0 kg)(9.80 m/s 2 )(0.200 m) = 114 J . ( b ) The potential energy on the second step is U 2 = mg 2 h = 2 U 1 = 2(114 J) = 228 J . ( c ) The potential energy on the third step is U 3 = mg 3 h = 3 U 1 = 3(114 J) = 342 J . ( d ) The potential energy on the n th step is U n = mgnh = nU 1 = n (114 J) = 114 n J . ( e ) The change in energy is ? E = U 2 U 6 = (2 – 6)(114 J) = – 456 J . 6. We use a body temperature of 98°F = 37°C = 310 K. We find the peak wavelength from P = (2.90 × 10 –3 m · K)/ T = (2.90 × 10 –3 m · K)/(310 K) = 9.4 × 10 –6 m = 9.4 m .

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Chapter 38 p. 2 7. ( a ) To find the wavelength when I ( λ , T ) is maximal at constant temperature, we set d I /d = 0: d I d = d d 2 π hc 2 –5 e hc / kT –1 =2 π hc 2 –6 e hc / kT hc / 2 kT e hc / kT e hc / kT 2 =– 2 π hc 2 6 5–( hc / kt ) e hc / kT e hc / kT – 1 2 = 0,which gives hc / P kT ) e hc / P kT –5=0, or 5–( hc / P kT )=5 e hc / P kT . This equation will have a solution P T = constant, which is the Wien displacement law. ( b ) To find the value of the constant, we let x = hc / P kT , so the transcendental equation is 5 x = 5 e x . One way to solve this equation is to plot each side against x . We see from the plot that the solution is very close to x = 5. If we let ? = 5 – x , we get ? = 5 e ( ? – 5) ˜ 5 e – 5 = 0.034, so x = 4.966. T h u s w e h a v e P T = hc / xk ; 2.90 × 10 –3 m · K = h (3.00 × 10 8 m/s)/(4.966)(1.38 × 10 –23 J/K), which gives h = 6.63 × 10 –34 J · s . ( c ) For the rate at which energy is radiated per unit surface area for all wavelengths we have I ( , T )d 0 = 2 π hc 2 e hc / kT d 0 .
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch38Word - Chapter 38 p. 1 CHAPTER 38 Early Quantum Theory...

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