Question

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

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

$8.04\times {10}^{-22}J$.

Step-by-step solution

Given data

k = 20 N/m

M = 0.207 kg

Solution

Mass of the one atom is,

$\mathrm{m}=\frac{\mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}}$

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

lead is,

$$\begin{gathered} \mathrm{E}=\hslash \mathrm{\omega }\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ =\frac{\mathrm{h}}{2\mathrm{r}}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ =\frac{\mathrm{h}}{2\mathrm{r}}\sqrt{\frac{\mathrm{k}}{\mathrm{M}/{\mathrm{N}}_{\mathrm{A}}}}\hspace{0.33em} \\ =\frac{6.626\times {10}^{-34}}{2(3.14)}\sqrt{\frac{20}{0.207/6.022\times {10}^{23}}} \\ =8.04\times {10}^{-22}\mathrm{J} \end{gathered}$$

Hence,

$\boxed{\mathrm{E}{=8.04\times 10}^{-22}\mathrm{J}}$