Question

The product of uncertainties in particle's momentum and position.

Short answer

The product of uncertainty in position and momentum is according to the uncertainty principle.

Step by Step Solution

The complete rule stipulates that the product of the uncertainties in position and velocity is Planck's constant, or about$6.6\times {10}^{-34}$ joule-second.

Step-by-step solution

The concept and the formula used.

The uncertainty in position is$L\sqrt{\frac{1}{12}-\frac{1}{2n^2{\pi }^2}}$ .

The uncertainty in momentum is $\frac{nhx}{L}$.

Formula used:

The uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives uncertainty in position if we have uncertainty in momentum.

Write the expression for the uncertainty principle.

$\Delta x\Delta p⩾\frac{\mathbb{A}}{2}$

Here, $\Delta x$is the change in position,$\Delta p$ is the change in momentum and h is the reduced Planck's constant.

Calculating the value using the formul

Write the expression for uncertainty in position.

$\Delta x=L\sqrt{\frac{1}{12}-\frac{1}{2n^2{x}^2}}\dots ...\text{(1)}$

Here,$\Delta x$is the uncertainty in position, L is the length and n is the number of state.

Write the expression for uncertainty in momentum.

$\Delta p=\frac{n\hat{\lambda }\pi }{L}\dots ..\text{(2)}$

Here,$\Delta p$is the uncertainty in momentum, n is the number of orbit and L is length.

Multiply equations (1) and (2).

$\Delta x\Delta p=n\pi \hslash \sqrt{\frac{1}{12}-\frac{1}{2n^2{\pi }^2}}$

$=\hslash \sqrt{\frac{n^2{\pi }^2}{12}-\frac{1}{2}}⩾\frac{\hslash }{2}$

For ground state, $n=1$the product of uncertainties is bigger than$\frac{n}{2}$, that means$\Delta x\Delta p⩾\frac{n}{2}$which is according to the uncertainty principle for position and momentum operator.

Thus, the product of uncertainty in position and momentum is according to uncertainty principle.