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?

Explanation / Answer

we know that

m = 0.027/6.02e23 = 4.4828 * 10 ^ -26

E=hbar*sqrt(64/m)

E=1.05457148e-34 * sqrt ( 64 / 4.4828 * 10 ^ -26 )

E = 3.98 * 10 ^ -21 JOULES ===============ANSWER)

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