Chapter 29, Problem 11P

(a)

To determine

The stiffness constant.

Answer to Problem 11P

The stiffness constant is $k=680.5\text{ N/m}$

Explanation of Solution

Given:

The given graph is shown below.

  

  $r_0=0.074\text{ nm}$

  $PE=-4.5\text{ eV}$

  $1\text{ eV}=1.6\times {10}^{-19}\text{J}$

Formula used:

Potential energy is given by: $PE=\frac{1}{2}k{x}^2$

Calculation:

First, calculate the value of $x$ as follows:

  $\begin{array}{l}x=r-r_0\\ x=\left(0.12\text{ nm}\right)-\left(0.074\text{ nm}\right)\\ x=0.046\text{ nm}\end{array}$

Rearranging the above given formula,

  $\begin{array}{l}k=\frac{2PE}{x^2}\\ k=\frac{2\left(4.5\text{ eV}\right)}{{\left(0.046\text{ nm}\right)}^2}\\ k=\frac{2\left(4.5\times 1.6\times {10}^{-19}\text{J}\right)}{{\left(0.046\times {10}^{-9}\text{m}\right)}^2}\\ k=680.5\text{ N/m}\end{array}$

Conclusion:

Thus, the stiffness constant is $k=680.5\text{ N/m}$

(b)

To determine

The fundamental wavelength for vibrational transitions.

Answer to Problem 11P

Fundamental wavelength is $\lambda =2.1\times {10}^{-6}\text{m}$

Explanation of Solution

Given:

Mass of proton is $m=1.67\times {10}^{-27}\text{kg}$

Speed of light is $c=3\times {10}^8\text{m/s}$

Formula used:

Wavelength can be given as:

  $\lambda =\frac{c}{f}$

Calculation:

Angular frequency can be written as:

  $\begin{array}{l}\omega =2\pi f\\ \sqrt{\frac{k}{m}}=2\pi f\\ f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\end{array}$

Now, using the above formula of wavelength,

  $\begin{array}{l}\lambda =\frac{c}{\left(\frac{1}{2\pi }\sqrt{\frac{k}{m}}\right)}\\ \lambda =2\pi c\sqrt{\frac{m}{k}}\end{array}$

Substituting all the values and also use half of the mass as given. So

  $\begin{array}{l}\lambda =2\left(3.14\right)\left(3\times {10}^8\text{m/s}\right)\sqrt{\frac{\frac{1}{2}\left(1.67\times {10}^{-27}\text{kg}\right)}{\left(680.5\text{ N/m}\right)}}\\ \lambda =2.1\times {10}^{-6}\text{m}\end{array}$

Conclusion:

Thus, the fundamental wavelength is $\lambda =2.1\times {10}^{-6}\text{m}$