91影视

Gaussian function of the form$蠄\left(x\right)伪e^{-\left(x^2/2{蔚}^2\right)}$is the standard deviation or uncertainty in position.

The probability density would be proportional to$蠄\left(x\right)$squared:$e^{-\left(x^2/2{蔚}^2\right)}$.

The uncertainty in position of the time-evolving Gaussian wave function of a free particle is given by

$螖x=蔚\sqrt{1+\frac{h^2{t}^2}{4m^2{蔚}^4}}$

$\begin{array}{rcl}蔚& =& 500鈥�\text{nm}\\ & =& (500鈥�\text{nm})\left(\frac{{10}^{-9}鈥�\text{m}}{\text{n}m}\right)\\ & =& 5脳{10}^{-7}鈥�\text{m}\\ & & \end{array}$

The formula for uncertainty in the position from the previous step can be rewritten in the following form.

$t=\sqrt{\left[{\left(\frac{螖x}{蔚}\right)}^2-1\right]\frac{4m^2{蔚}^4}{h^2}}$ ... (1)

Where,

$蔚$is standard deviation or uncertainty in position, is the mass of an electron $9.11脳{10}^{-31}鈥�\text{kg}$, t is the time require for an electron to reach the length of an automobile$螖x$, h is the Planck constant$1.054脳{10}^{-34}鈥�\text{J.s}$

Substitute the values from previous step,

$5脳{10}^{-7}鈥�\text{m}$for$蔚$ , $9.11脳{10}^{-31}鈥�\text{kg}$for $m$, 5m for$螖x$ in equation (1) to solve t

$\begin{array}{rcl}t& =& \sqrt{\left[{\left(\frac{螖x}{蔚}\right)}^2-1\right]\frac{4m^2{蔚}^4}{h^2}}\\ & =& \sqrt{\left[{\left(\frac{5鈥�\text{m}}{\left(5脳{10}^{-7}鈥�\text{m}\right)}\right)}^2-1\right]\frac{4{\left(9.11路{10}^{-31}鈥�\text{kg}\right)}^2{\left(5脳{10}^{-7}鈥�\text{m}\right)}^4}{{\left(1.054脳{10}^{-34}鈥�\text{J.s}\right)}^2}}\\ & =& 0.043鈥�\text{s}\\ & & \end{array}$

Therefore, the time required to reach length of an automobile is $0.043鈥�\text{s}$