In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of aluminum to be 16 N/m. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective "interatomic spring stiffness" for an oscillator is 4*16 N/m = 64 N/m. The mass of one mole of aluminum is 27 grams (0.027 kilograms). What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of aluminum? one quantum = 6.619566e-45 joules
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Given: Mass of 1 mole of aluminum = 0.027 kg Avogadro's number = \(6.022 \times 10^{23}\) Mass of one atom = Mass of 1 mole / Avogadro's number Mass of one atom = 0.027 kg / \(6.022 \times 10^{23}\) Mass of one atom ≈ \(4.48 \times 10^{-26}\) kg Show more…
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