91Ó°ÊÓ

The given data can be listed below,

• The mass of the particle is,$m=1\text{‰Ӽg}$
• The size of the enclosure is, $L=1\text{ c³¾}$
• The taken to complete bounce back is,$T=2\text{ y±ð²¹°ù}\left(\frac{3.156×{10}^7\text{ s}}{1\text{ y±ð²¹°ù}}\right)$

When the system is bounded in the conventional sense, a stationary state is said to be a bound state. In classical mechanics, a system with some finite energy cannot exist in a region of infinite potential energy in an infinite potential well.

It should be noted that the energy quantization only takes place when the system is bounded.

The energy of the bound state is given by,

$E=\frac{n{h}^2}{8m{L}^2}$ …(¾±)

Here, nis the number nodes or state, h is the plank’s constant whose value is $\mathrm{ν}$$6.63×{10}^{−34}\text{ J}â‹\dots \text{s}$, m is the mass of the particle and L is the length of enclosure.

According to Einstein, the photon energy of the particle is given by,

$E=h\mathrm{ν}$ …(¾±¾±)

Here, $\mathrm{ν}$is the frequency of the particle.

Compare two equations the number of nodes is given by,

$\begin{array}{c}h\mathrm{ν}=\frac{n{h}^2}{8m{L}^2}\\ n=\frac{8m{L}^2}{hT}\end{array}$

Substitute all the values in the above,

$\begin{array}{c}n=\frac{8\left(1\text{‰Ӽg}\right){\left(1\text{ c³¾}\right)}^2}{(6.63×{10}^{−34}\text{ J}â‹\dots \text{s})(6.037×{10}^7\text{ s})}\\ =1.99×{10}^{13}\text{ n´Ç»å±ð²õ}\\ ≈\text{2}×{\text{10}}^{13}\text{ n´Ç»å±ð}\end{array}$

Thus, the nodes in the enclosure are $\text{2}×{\text{10}}^{13}\text{ n´Ç»å±ð²õ}$.

The momentum would be greater if the particle moved more quickly or had more mass, which would suggest a shorter wavelength and, as a result, a greater number of nodes. It would never be anticipated to act as a wave since it would be more massive or travelling faster, and more mass.

Thus, when the particle is more massive or moving faster the momentum of the particle is increased resulting in shorter wavelength and the nodes will also increase