Quantum energy using spring stiffness

[firegoalie33]

Homework Statement

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?
one quantum = ? Joules.

Homework Equations

I have in my book the following "energy can be added to a one-dimensional atomic oscillator only in multiples of one "quantum" of energy (h_bar*omega_naught) = sqrt(ks,i/ma) where h_bar = h/2pi = 1.05e-34 joule*second. ks,i is the interatomic spring stiffness and ma is the mass of the atom.

The Attempt at a Solution

I tried this problem by doing 1.05e-34*(sqrt(20/.207)) but this was wrong.

Any help would be greatly appreciated.
thanks!

 

[anathema]

The correct way to approach this problem is to use the formula provided in your book: E = h_bar*omega_naught = sqrt(ks,i/ma). In this case, h_bar is a constant, so we can simply focus on the square root term.

First, we need to calculate the value of ks,i, the interatomic spring stiffness for an oscillator in a block of lead. Since each atom is connected to two springs, the effective stiffness is 4*5 N/m = 20 N/m.

Next, we need to determine the mass of the atom, ma. We are given the mass of one mole of lead (207 grams), so we can use the molar mass of lead (207 g/mol) to calculate the mass of one atom: 207 g/mol / 6.022 x 10^23 atoms/mol = 3.44 x 10^-23 grams.

Now, we can plug these values into the formula: E = sqrt(20 N/m * 3.44 x 10^-23 g) = 4.56 x 10^-12 joules.

Therefore, the energy of one quantum for an atomic oscillator in a block of lead is 4.56 x 10^-12 joules.

I hope this helps! Let me know if you have any further questions.