Question

In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead to be 5 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*5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms).
What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?

Answer 1

Answer:

8.01e-22

Explanation:

Answer 2

The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

Extension produced by one mole of lead

The extension produced by one mole of lead atom is calculated by applying Hooke's law;

F = kx

mg = kx

x = mg/k

x = (0.207 x 9.8) / (20)

x = 0.101 m

Energy stored in the lead block

The Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;

E = ¹/₂kx²

E = ¹/₂ (20)(0.101)²

E = 0.102 J

Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

Learn more about Hooke's law here: https://brainly.com/question/2648431