In Chapter 4 you determined the stiffness of the interatomic “spring” (chemical bond) between atoms in a block of lead to be 5 N/m, based on the value of Young’s modulus for lead. 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 × 5 N/m = 20 N/m. The mass of one mole of lead is 207 g (0.207 kg). What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?

Short Answer

Expert verified

The energy of one quantum of energy for an atomic oscillator in a block of lead is

8.04×10-22J.

Step by step solution

01

Given data

k = 20 N/m

M = 0.207 kg

02

Solution

Mass of the one atom is,

m=MNA

The energy of one quantum of energy for an atomic oscillator in a block of

lead is,

E=ω=h2rkm=h2rkM/NA=6.626×10-342(3.14)200.207/6.022×1023=8.04×10-22J

Hence,

E=8.04×10-22J

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