91影视

Potential energy of simple harmonic oscillator is given by,

$U=\frac{1}{2}K{x}^2$

Here, k represents spring constant and x represents displacement

The lowest energy is given by,

$\text{E=}\frac{\text{h}}{\text{2}}\sqrt{\frac{\text{k}}{\text{m}}}$

Here, h represents Planck鈥檚 constant and m represents mass of particle.

When the potential energy is more than the total energy, the region becomes classically forbidden,

$\begin{array}{l}\frac{1}{2}k{x}^2=\frac{h}{2}\sqrt{\frac{k}{m}}\\ x=卤\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}\end{array}$

But, $b={\left(\frac{mk}{h^2}\right)}^{1/4}$

The classically forbidden region lies from$鈭�\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}$ to$\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}$ .

The wave function is given by:

${蠄}_0\left(x\right)={\left(\frac{b}{\sqrt{蟺}}\right)}^{1/2}{e}^{鈭�b^2{x}^2/2}$

The probability to find the particle in the forbidden region,

${鈭�}_{\frac{\sqrt{\text{A}}}{{\text{(km)}}^{\text{1/4}}}}^{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{1/4}}}}{\text{蠄(虫)蠄}}^{\text{0}}\text{(x)dx}$

By symmetry the above function is changed into,

$\begin{array}{l}{鈭�}_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{14}}}}^{{\text{(km)}}^{\text{1/4}}}{\text{蠄(虫)蠄}}^{\text{0}}\text{(x)dx}\\ \text{=2}{鈭�}_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{14}}}}^{\text{楼}}{\left(\sqrt{\frac{\text{b}}{\sqrt{\text{蟺}}}}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}\text{/2}}\right)}^{\text{2}}\text{dx}\\ \text{=2}\frac{\text{b}}{\sqrt{\text{蟺}}}{鈭�}_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{1/4}}}}^{\text{楼}}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}}\text{dx}\\ \text{=}\frac{\text{2}}{\sqrt{\text{蟺}}}\underset{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{3/4}}}}{\overset{\text{楼}}{鈭�}}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}}\text{bdx}\end{array}$

By change of variable method, substitute$y=bx$and$dy=bdx$.

$\begin{array}{l}\frac{2}{\sqrt{蟺}}{鈭�}_{\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}}^{\mathrm{鈭�}}{e}^{鈭�b^2{x}^2}bdx\\ =\frac{2}{\sqrt{蟺}}{鈭�}_{\frac{b\sqrt{\mathrm{鈩�}}}{{\left(km\right)}^{1/4}}}^{\mathrm{鈭�}}{e}^{鈭�y^2}dy\end{array}$

Since, $b={\left(\frac{mk}{h^2}\right)}^{1/4}$above integral becomes,

$\begin{array}{l}\frac{2}{\sqrt{蟺}}\underset{1}{\overset{\mathrm{鈭�}}{鈭�}}{e}^{鈭�y^2}dy=\frac{2}{\sqrt{蟺}}\left(0.1394\right)\\ =0.157299\end{array}$

Therefore,

The required probability of finding the particle is 0.1573