Chapter 28, Problem 10P

To determine

The uncertainty in position of electron and basketball.

Answer to Problem 10P

The uncertainty in the momentum is $\text{1}\text{.055}\text{×}{\text{10}}^{\text{-19}}\text{kg}\cdot \text{m/s}$

The lowest possible kinetic energy $\text{ 21}\text{MeV}$

Explanation of Solution

Given info: Radius is $\text{1}\text{.0}{\text{×10}}^{\text{-15}}\text{m}$

Formula used:

It is known that $\text{Δp}\ge \frac{\text{h}}{\text{Δx}}$

Where,

  $h$ is Planck’s constant, $1.055\times {10}^{-34}J\cdot s$

  $\Delta \text{p}$ is uncertainty in the momentum

  $\Delta x$ is the uncertainty in the position measurement

Calculation:

Use the radius as the uncertainty in position for the neutron to calculate the uncertainty in the momentum by $\text{Δp}\ge \frac{\text{h}}{\text{Δx}}$

  $\begin{array}{l}\text{=}\frac{\left(\text{1}\text{.055}\text{×}{\text{10}}^{\text{-34}}\text{Jgs}\right)}{\left(\text{1}{\text{.0×10}}^{\text{-15}}\text{m}\right)}\\ \text{=1}\text{.055}\text{×}{\text{10}}^{\text{-19}}\text{kg}\cdot \text{m/s}\end{array}$

Suppose that the lowest value for momentum is the least uncertainty. So, estimate the lowest possible kinetic energy as

  $\begin{array}{l}\text{E =}\frac{{\left(\text{Δp}\right)}^{\text{2}}}{\text{2m}}\text{=}\frac{{\left(\text{1}{\text{.055×10}}^{\text{-19}}\text{kg×m/s}\right)}^{\text{2}}}{\text{2}\left(\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}\right)}\\ \text{= 3}{\text{.33×10}}^{\text{-12}}\text{J }\text{= 21}\text{MeV}\end{array}$

Conclusion:

All the required values arecalculated.