Chapter 28, Problem 11P

To determine

The lowest possible energy.

Answer to Problem 11P

The answer is $\text{200}\text{MeV}$

Explanation of Solution

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

Mass of an electron $\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}$

Formula used:

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

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 electron.

Now calculate the uncertainty in momentum by using the formula $\Delta p\ge \frac{h}{\Delta \text{x}}$

  $\text{Δp}\ge \frac{\text{h}}{\text{Δx}}\text{=}\frac{\left(\text{1}{\text{.055×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)}{\left(\text{1}{\text{.0×10}}^{\text{-15}}\text{m}\right)}=1.055\times {10}^{-19}\text{kg}\cdot \text{m/s}$

Suppose that the lowest value for the momentum is the least uncertainty. So, calculate the lowest possible energy by using the formula $\text{E=}\left(\text{KE}\right){\text{+m}}_{\text{0}}{\text{c}}^{\text{2}}\mathrm{...}\left(\text{1}\right)$

Expand the equation $\left(\text{1}\right)$

  $\begin{array}{l}\text{E=}\left(\text{KE}\right)\text{+}{\text{m}}_{\text{0}}{\text{c}}^{\text{2}}\\ \text{=}{\left({\text{p}}^{\text{2}}{\text{c}}^{\text{2}}{\text{+m}}_{\text{0}}{\text{c}}^{\text{4}}\right)}^{\frac{\text{1}}{\text{2}}}\\ \text{=}{\left[{\left(\text{Δp}\right)}^{\text{2}}{\text{c}}^{\text{2}}{\text{+m}}_{\text{0}}{}^{\text{2}}{\text{c}}^{\text{4}}\right]}^{\frac{\text{1}}{\text{2}}}\mathrm{...}\left(2\right)\end{array}$

Where,

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

  $\text{c}$ is speed of light, $\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}$

  ${\text{m}}_{\text{0}}$ is rest mass,

Substitute all the known and given values in equation $\left(\text{2}\right)$

  $\text{=}{\left[{\left(\text{1}{\text{.055×10}}^{\text{-19}}\text{kg×m/s}\right)}^{\text{2}}{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\text{+}{\left(\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)}^{\text{2}}{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}^{\text{4}}\right]}^{\frac{\text{1}}{\text{2}}}$

  $\text{=3}{\text{.175×10}}^{\text{-11}}\text{J}\approx \text{200}\text{MeV}$

Conclusion:

The required value is $\text{200}\text{MeV}$ .