Question

One of the cornerstones of quantum mechanics is that bound particles cannot be stationary-even at zero absolute temperature! A "bound" particle is one that is confined in some finite region of space. as is an atom in a solid. There is a nonzero lower limit on the kinetic energy of such a particle. Suppose minimum kinetic energy of width $\mathbf{L}$. Obtain an approximate formula for its minimum kinetic energy.

Short answer

The minimum kinetic energy formula is$K·E.\ge \frac{h^2}{8m{L}^2}$

Step-by-step solution

Uncertainty principle.

That the position and the velocity of an object cannot both be measured exactly, at the same time, even in theory.

$\Delta x∆p\ge \frac{\mathrm{h}}{4\mathrm{\pi }}$

$\Delta x$uncertainty in position.

$\Delta p$uncertainty of momentum.

$h$ Planck's constant.

Kinetic energy.

$\Delta x\Delta p=\Delta x\Delta \left(mv\right)⩾\frac{h}{2}$

localid="1659187305509" $\Delta v=\frac{h}{2m\Delta x}$

Here $\Delta \mathrm{x}=\mathrm{L}$

$\Delta v=\frac{h}{2mL}$

The kinetic energy$K.E.=\frac{1}{2}m(\Delta v{)}^2$

Substitute the value,

localid="1659187247106" $\text{K.E.≥}\frac{{\text{h}}^{\text{2}}}{{\text{8mL}}^{\text{2}}}$

Therefore, the minimum Kinetic Energy formula of the width is localid="1659187237799" $\text{K.E≥.}\frac{{\text{h}}^{\text{2}}}{{\text{8mL}}^{\text{2}}}$