91影视

The quantized vibrational energy levels for an atomic harmonic oscillator are given by,

$E_N=Nh{蝇}_0+E_0$

Here,

$N$ is the principal quantum number.

$E_0$ is the ground state energy of harmonic oscillator.

${蝇}_0$ is the angular frequency.

$h$ is Planck鈥檚 constant.

$k_s$ is interatomic spring stiffness.

$m$ is mass of an atom.

So the energy of a spring-mass oscillator is given by,

$E=\frac{1}{2}{k}_s{A}^2$

where $A$ is amplitude of oscillation.

On the equating the equation $E_N=Nh{蝇}_0+E_0$ and$E=\frac{1}{2}{k}_s{A}^2$ . We get,

$\begin{array}{rcl}Nh{蝇}_0+E_0& =& \frac{1}{2}{k}_s{A}^2\\ Nh{蝇}_0+\frac{1}{2}h{蝇}_0& =& \frac{1}{2}{k}_s{A}^2\\ \left(N+\frac{1}{2}\right)h{蝇}_0& =& \frac{1}{2}{k}_s{A}^2\\ \left(N+\frac{1}{2}\right)h\sqrt{\frac{k_s}{m}}& =& \frac{1}{2}{k}_s{A}^2\\ & & \end{array}$

By simplifying we get,

$\begin{array}{rcl}N+\frac{1}{2}& =& \frac{\frac{1}{2}{k}_s{A}^2}{\left(h\sqrt{\frac{k_s}{m}}\right)}\\ N& =& \frac{\frac{1}{2}{k}_s{A}^2}{\left(h\sqrt{\frac{k_s}{m}}\right)}-\frac{1}{2}\\ & & \end{array}$

On substituting the known values on the above equation. We get,

$\begin{array}{rcl}N& =& \frac{\frac{1}{2}\left(0.7\right){\left(0.2\right)}^2}{\left(\left(1路05脳{10}^{-34}\right)\sqrt{\frac{0.7}{0.02}}\right)}-\frac{1}{2}\\ & =& 2脳{10}^{31}\\ & & \end{array}$

The number of levels above the ground state of the spring oscillator is $2脳{10}^{31}$