Chapter 6, Problem 116AP

Interpretation Introduction

Interpretation:

The uncertainty in the position of the electron with given speed is to be determined, a comment on the result is to be stated, and the uncertainty in the position of the ball is to be determined.

Concept introduction:

Heisenberg gave the uncertainty principle, which statethat the product of the uncertainty in position and momentum of a particle cannot be less than $\frac{\text{h}}{4\pi }$.

It is represented as follows:

$\Delta x\cdot \Delta p\ge \frac{\text{h}}{4\pi }$

Here, $h$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, and the uncertainty in position and momentum is $\Delta x$ and $\Delta p$, respectively.

As momentum is the product of mass and velocity, the equation of Heisenberg uncertainty principle can also be represented as shown below:

$\Delta x\cdot m\Delta u\ge \frac{\text{h}}{4\pi }$

Here, $\Delta x$ is uncertainty in the position of a particle, $m$ is the mass of the particle, $\Delta u$ is uncertainty in velocity, and $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$.

The minimum uncertainty in speed is calculated by using the following relation:

$\Delta u=\text{Speed}\times \text{Uncertainty of speed}$

The uncertainty in momentum can be evaluated as

$\Delta p=\text{momentum}\times \text{Uncertainty of momentum}$

The relationship between $\text{J}$ and ${\text{kg m}}^2{\text{ s}}^{-2}$ can be expressed as $1\text{J}\text{=}1{\text{ kg m}}^2{\text{ s}}^{-2}$.

The conversion factor is $\frac{1{\text{ kg m}}^2{\text{ s}}^{-2}}{1\text{ J}}$.

Answer to Problem 116AP

Solution:

(a) ${\text{6×10}}^{-11}\text{ m/s}$

(b) $7.9{\text{×10}}^{-29}\text{ m}$

Explanation of Solution

a)The minimum uncertainty in an electron’s position

The mass of the electron is $9.1094\times {10}^{-31}\text{ kg}$.

The radius of a hydrogen atom is $5.29\times {10}^{-11}\text{ m}$.

The speed of the electron is $5\times {10}^6\text{ m/s}$.

The uncertainty in speed can be evaluated as

$\Delta u=\text{Speed}\times \text{Uncertainty of speed}$

$\begin{array}{c}\Delta u=5\times {10}^6\text{ m/s}\times \left(0.2\right)\\ =1\times {10}^6\text{ m/s}\end{array}$

So, the uncertainty in the speed of the electronis $1\times {10}^6\text{ m/s}$.

The uncertainty in the position of theelectron can be evaluated as

$\begin{array}{c}\Delta x\left(9.1094\times {10}^{-31}\text{ kg}\right)\left(1\times {10}^6\text{ m/s}\right)=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi }\\ \Delta x=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi \left(9.1094\times {10}^{-31}\text{ kg}\right)\left(1\times {10}^6\text{ m/s}\right)}\left(\frac{1{\text{ kg m}}^2{\text{ s}}^{-2}}{1\text{ J}}\right)\\ =5.8{\text{×10}}^{-11}\text{ m/s}\\ ={\text{6×10}}^{-11}\text{ m/s}\end{array}$

Therefore, the uncertainty in the position of the electronis ${\text{6×10}}^{-11}\text{ m/s}$.

Since the value for $\Delta x$ of electron is greater than the radius of the hydrogen atom, it is hard to predict the exact location of the electron in the atom.

The momentum of $0.15\text{ kg}$ ball is $\text{6}\text{.7 kg}\cdot \text{m/s}$.

b) Uncertainty in baseball’s position.

The momentum of $0.15\text{ kg}$ ball is $\text{6}\text{.7 kg}\cdot \text{m/s}$.

The uncertainty in momentum can be evaluated as

$\Delta p=\text{momentum}\times \text{Uncertainty of momentum}$

$\begin{array}{c}\Delta p=\left(1.0\times {10}^{-7}\right)\left(\text{6}\text{.7 kg}\cdot \text{m/s}\right)\\ =\text{6}\text{.7}\times {10}^{-7}\text{kg}\cdot \text{m/s}\end{array}$

So, the uncertainty in the momentum of the baseball is $\text{6}\text{.7}\times {10}^{-7}\text{kg}\cdot \text{m/s}$.

The value for $\Delta x$ of the baseball can be evaluated as

$\begin{array}{c}\Delta x\cdot \Delta p=\frac{h}{4\pi }\\ \Delta x\left(\text{6}\text{.7}\times {10}^{-7}\text{kg}\cdot \text{m/s}\right)=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi }\\ \Delta x=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi \left(\text{6}\text{.7}\times {10}^{-7}\text{kg}\cdot \text{m/s}\right)}\left(\frac{1{\text{ kg m}}^2{\text{ s}}^{-2}}{1\text{ J}}\right)\\ =7.9{\text{×10}}^{-29}\text{ m}\end{array}$

Therefore, the value for $\Delta x$ of the baseball is $7.9{\text{×10}}^{-29}\text{ m}$.