Question

What is the product of uncertainties determined in Exercise 60 and 61? Explain.

Short answer

The product of uncertainty in position and momentum is $\frac{\hslash }{2}$, which is in accordance with the uncertainty principle.

Step-by-step solution

The concepts and formulas used to solve the given problem.

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.

The expression for the uncertainty principle.

$\Delta x\Delta p⩾\frac{h}{2}$

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

$\Delta x=\frac{1}{\sqrt{2}}{\left(\frac{h^2}{mk}\right)}^{1/4}$ …… (1)

Here, $∆x$is the uncertainty in position, h is reduced Planck's constant, m is the mass and k is the constant.

The expression for uncertainty in momentum.

$\Delta p=\left(\sqrt{\frac{h}{2}}\right)(mk{)}^{1/4}$ …… (2)

Here, $∆p$is the uncertainty in momentum, $\hslash$is reduced Planck's constant, m is the mass and k is the constant.

Step 2:Product

Multiply equation (1) and (2).

$\Delta x\Delta p=\frac{1}{\sqrt{2}}{\left(\frac{h^2}{mk}\right)}^{1/4}\left(\sqrt{\frac{h}{2}}\right)(mk{)}^{1/4}$

$=\frac{h}{2}$

The product of uncertainties is equal to $\frac{h}{2}$which is in accordance with the Uncertainty principle.