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Mass of an atom of neon-22, m = 21.9914u = 3.82 X 10-26 kg In quantum physics, the de Broglie wavelength is defined as 入 = h / p where 入 is the de Broglie wavelength h is Planck constant, h = 6.63 X 10-34 Js p is the momentum of the particle So, 3.27 X 10-12 = 6.63 X 10-34 / p Momentum, p = 2.03 X 10-22 kgms-1 Momentum, p = mv So, speed of the atom, v = 2.03 X 10-22 / 3.82 X 10-26 = 5.31 X 103 ms-2 As this speed is much lower than speed of light, so the relativistic effect can be ignored. K.E. of the atom = 1/2 mv2 = 1/2 (3.82 X 10-26)(5.31 X 103)2 = 5.38 X 10-19 J = 3.36 eV
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What would be the kinetic energy of an atom of neon-22 (molar mass = 21.9914 g/mol) if its de Broglie wavelength were 3.27 picometers?最佳解答:
Mass of an atom of neon-22, m = 21.9914u = 3.82 X 10-26 kg In quantum physics, the de Broglie wavelength is defined as 入 = h / p where 入 is the de Broglie wavelength h is Planck constant, h = 6.63 X 10-34 Js p is the momentum of the particle So, 3.27 X 10-12 = 6.63 X 10-34 / p Momentum, p = 2.03 X 10-22 kgms-1 Momentum, p = mv So, speed of the atom, v = 2.03 X 10-22 / 3.82 X 10-26 = 5.31 X 103 ms-2 As this speed is much lower than speed of light, so the relativistic effect can be ignored. K.E. of the atom = 1/2 mv2 = 1/2 (3.82 X 10-26)(5.31 X 103)2 = 5.38 X 10-19 J = 3.36 eV
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