In an earlier chapter you calculated the stiffness of the interatomic \"spring\"

Question

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 = ??? joules

Explanation / Answer

mass =0.027/6.022*1023 =4.48*10-26

E= hbar*(K/m)1/2 =1.0545*10-34(64/4.48*10-26)1/2 =4*10-21 J

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