# Model assume the blackbodies obey weins law in

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Model: Assume the blackbodies obey Wein’s law in Equation 38.9: 6 peak (2 90 10 nm K)/ . T λ = . × Visualize: We want to solve for T in the equation above. We are given peak 300 nm 3000 nm. λ = , Solve: 6 peak 2 90 10 nm K T λ . × = (a) 6 2 90 10 nm K 9667K 9394 C 300nm T . × = = = ° (b) 6 2 90 10 nm K 966 7K 694 C 3000nm T . × = = . = °

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38.24. Model: Assume the metal sphere is a blackbody (so the emissivity e = 1 ). Visualize: First use Wein’s law (Equation 38.9) to find the temperature, then use Stefan’s law (Equation 38.8) to determine the power radiated. We are given 1 0 cm R = . and peak 2000 nm. λ = Solve: 6 2 90 10 nm K 1450 K 2000 nm T . × = = 4 8 2 4 2 4 (1)(5 67 10 W/m K )[4 (1 0 cm) ](1450 K) 315 W Q e AT t σ π = = . × . = Δ Assess: The sphere radiates more than 3100W light bulbs, but it has a larger surface area than the filaments, so the answer is reasonable.
38.25. Model: Assume the ceramic cube is a blackbody (so the emissivity e = 1 ). Visualize: First use Stefan’s law (Equation 38.8), 4 Q t e AT σ Δ = , to find the temperature, then use Wein’s law (Equation 38.9) to get the peak wavelength. We are given 2 6(3 0 cm 3 0 cm) 0 0054 m A = . × . = . and 630 W. Q t /Δ = Solve: Solve Stefan’s law for T . 4 4 8 2 4 2 630 W 1198 K (1)(5 67 10 W/m K )(0 0054 m ) Q t T e A σ = = = . × . Now plug this temperature into Wein’s law. 6 6 peak 2 90 10 nm K 2 90 10 nm K 2420 nm 2 42 m 1198 K T λ μ . × . × = = = = .

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38.26. Model: Use the relativistic expression for the total energy. Solve: (a) The energy of the proton is ( ) ( ) ( )( ) 2 2 2 2 27 8 p 2 2 2 9 9 19 1 1 7.089 1.67 10 kg 3.0 10 m/s 1 1 0.99 1 eV 1.065 10 J 6.66 10 eV 6660 MeV 1.6 10 J E mc mc mc v c γ = = = = × × = × × = × = × (b) Likewise, the energy of the electron is ( ) ( )( ) 2 31 8 13 6 19 2 1 1 eV 9.11 10 kg 3.0 10 m/s 5.812 10 J 3.63 10 eV 3.63 MeV 1.6 10 J 1 0.99 E = × × = × × = × = × Assess: The total energy E for the proton is larger than that for the electron by the factor m proton / m electron = 1833.
38.27. Model: Use the relativistic expression for the total energy. Solve: (a) The energy of the proton is ( )( ) 2 27 8 19 2 9 p 2 2 2 2 3 1.67 10 kg 3.0 10 m/s 1.60 10 J 500 GeV 500 10 eV 1 eV 1 1 1.879 10 0.999998 E mc v c v c v c γ × × × = = = = × × = × = (b) Likewise for the electron, ( )( ) 2 31 8 2 2 9.11 10 kg 3.0 10 m/s 2.0 GeV 1 E v c × × = = 2 2 4 1 2.562 10 0.99999997 v c v c = × =

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38.28. Solve: The rest energy of an electron is ( )( ) 2 2 31 8 14 0 19 1 eV 9.11 10 kg 3.0 10 m/s 8.199 10 J 0.512 MeV 1.6 10 J E mc = = × × = × × = × The rest energy of a proton is ( )( ) 2 2 27 8 10 0 19 1 eV 1.67 10 kg 3.0 10 m/s 1.503 10 J 939 MeV 1.6 10 J E mc = = × × = × × = ×
38.29. Solve: (a) The kinetic energy of the electron is ( ) ( ) ( )( ) 2 2 31 8 15 19 1 eV 1 1.01 1 9.11 10 kg 3.0 10 m/s 0.8199 10 J 0.00512 MeV 1.6 10 J K mc γ = = × × = × × = × (b) Likewise for the proton, ( ) ( )( ) 2 27 8 1.01 1 1.67 10 kg 3.0 10 m/s 9.39 MeV K = × × = (c) Likewise for the alpha particle, ( )( ) ( )( ) 2 27 8 1.01 1 4 1.67 10 kg 3.0 10 m/s 37.6 MeV K = × × =

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38.30. Model: Mass and energy are equivalent. Solve: The energy released as kinetic energy is ( ) ( ) ( )( ) 2 2 27 8 11 19 1 eV 0.185 1.66 10 kg 3.0 10 m/s 2.76 10 J 173 MeV 1.6 10 J K E m c = Δ = Δ = × × = × × = ×
38.31.

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