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(B) The Heisenberg uncertainty principle states that $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. The lowest energy of a particle confined in a regio

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$IV < II < I < III$

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(B) The Heisenberg uncertainty principle states that $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. The lowest energy of a particle confined in a region of size $\Delta x$ is proportional to $\frac{1}{m(\Delta x)^2}$.$(IV)$ $A$ $H_{2}$ molecule in a nanotube is confined to a relatively large space (nanometer scale) and has a large mass,leading to the lowest energy.$(II)$ $A$ hydrogen atom in a $H_{2}$ molecule is confined by interatomic forces within the molecule,resulting in a smaller confinement region than the nanotube,thus higher energy than $(IV)$.$(I)$ An electron in a $H_{2}$ molecule is much lighter than an atom,and its confinement within the molecular orbital leads to a higher kinetic energy than the atom.$(III)$ $A$ proton in the carbon nucleus is confined to an extremely small region (femtometer scale) by strong nuclear forces,resulting in the highest energy due to the uncertainty principle.Therefore,the correct order of increasing lowest energy is $IV < II < I < III$.

Using the Heisenberg uncertainty principle,arrange the following particles in the order of increasing lowest energy possible.
$(I)$ An electron in $H_{2}$ molecule
$(II)$ $A$ hydrogen atom in a $H_{2}$ molecule
$(III)$ $A$ proton in the carbon nucleus
$(IV)$ $A$ $H_{2}$ molecule within a nanotube

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Using the Heisenberg uncertainty principle,arrange the following particles in the order of increasing lowest energy possible.$(I)$ An electron in $H_{2}$ molecule$(II)$ $A$ hydrogen atom in a $H_{2}$ molecule$(III)$ $A$ proton in the carbon nucleus$(IV)$ $A$ $H_{2}$ molecule within a nanotube