현대물리학 beiser 6판 3장 연습문제 영문판 Down
현대물리학 beiser 6판 3장 연습문제 영문판 Down
현대물리학 beiser 6판 3장 연습문제 영문판
현대물리학 beiser 6판 3장 연습문제 영문판
Chapter 3. Problem Solutions
1. A photon and a particle have the same wavelength. Can anything be said about how their linear momenta compare About how the photon`s energy compares with the particle`s total energy About how the photon’s energy compares with the particle`s kinetic energy
¡¼Sol¡½ From Equation (3.1), any particle’s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. With a common momentum p, the photon’s energy is pc, and the particle’s energy is ( pc )2 + (mc 2 )2 , which is necessarily greater than pc for a massive particle. The particle’s kinetic energy is K = E mc 2 =
(pc )2 + (mc 2 )
2
mc 2
For low values of p (p``mc for a nonrelativistic massive particle), the kinetic energy is K p2/2m, which is necessarily less than pc. For a relativistic massive particle, K pc mc2, and K is less than the photon energy. The kinetic energy of a massive particle will always be less than pc, as can be seen by using E = (pc)2 + (mc2)2 to obtain
( pc )2 K 2 = 2Kmc 2.
Inha University
Department of Physics
Chapter 3. Problem Solutions
3. Find the de Broglie wavelength of a 1.0-mg grain of sand blown by the wind at a speed of 20 m/s. ¡¼Sol¡½ For this nonrelativistic case, h 6.63 × 1034 J s λ= = = 3.3 × 10 29 m; 6 mv (1.0 × 10 kg)(20 m/s) quantum effects certainly would not be noticed for such an object. By what percentage will a nonrelativistle calculation of the de Broglie wavelength of a 100-keV electron be in error ¡¼Sol¡½ Because the de Broglie wavelength depends only on the electron`s momentum, the percentage error in the wavelength will be the same as the percentage error in the reciprocal of the momentum, with the nonrelativistic calculation giving the higher wavelength due to a lower calculated momentum. The nonrelativistic momentum is pnr = 2mK = 2( 9.1 × 1031 kg)(100 × 103 eV)(1.6 × 10-19 J/eV) = 1.71 × 10 22 kg m /s, and the relativistic momentum is pr = 1 c 5.
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=17041032&sid=knp868group1&key=
현대물리학 beiser 6판 3장 연습문제 영문판
현대물리학 beiser 6판 3장 연습문제 영문판
Chapter 3. Problem Solutions
1. A photon and a particle have the same wavelength. Can anything be said about how their linear momenta compare About how the photon`s energy compares with the particle`s total energy About how the photon’s energy compares with the particle`s kinetic energy
¡¼Sol¡½ From Equation (3.1), any particle’s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. With a common momentum p, the photon’s energy is pc, and the particle’s energy is ( pc )2 + (mc 2 )2 , which is necessarily greater than pc for a massive particle. The particle’s kinetic energy is K = E mc 2 =
(pc )2 + (mc 2 )
2
mc 2
For low values of p (p``mc for a nonrelativistic massive particle), the kinetic energy is K p2/2m, which is necessarily less than pc. For a relativistic massive particle, K pc mc2, and K is less than the photon energy. The kinetic energy of a massive particle will always be less than pc, as can be seen by using E = (pc)2 + (mc2)2 to obtain
( pc )2 K 2 = 2Kmc 2.
Inha University
Department of Physics
Chapter 3. Problem Solutions
3. Find the de Broglie wavelength of a 1.0-mg grain of sand blown by the wind at a speed of 20 m/s. ¡¼Sol¡½ For this nonrelativistic case, h 6.63 × 1034 J s λ= = = 3.3 × 10 29 m; 6 mv (1.0 × 10 kg)(20 m/s) quantum effects certainly would not be noticed for such an object. By what percentage will a nonrelativistle calculation of the de Broglie wavelength of a 100-keV electron be in error ¡¼Sol¡½ Because the de Broglie wavelength depends only on the electron`s momentum, the percentage error in the wavelength will be the same as the percentage error in the reciprocal of the momentum, with the nonrelativistic calculation giving the higher wavelength due to a lower calculated momentum. The nonrelativistic momentum is pnr = 2mK = 2( 9.1 × 1031 kg)(100 × 103 eV)(1.6 × 10-19 J/eV) = 1.71 × 10 22 kg m /s, and the relativistic momentum is pr = 1 c 5.
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=17041032&sid=knp868group1&key=
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