Chapter 28, Problem 45GP

(a)

To determine

The wavelength of the bullet.

Answer to Problem 45GP

  $\text{3}\text{.1}\text{×}{\text{10}}^{\text{-34}}\text{m}$

Explanation of Solution

Given info: The mass of bullet is $\text{12}\text{g}\text{=}\text{12}\text{×}{\text{10}}^{\text{-3}}\text{kg}$ .

Speed is $\text{180}\text{m/s}$ .

Formula used:

  $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}\mathrm{v}}$

Where, $\mathrm{h}$ is Planck’s constant, $6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$ , $\mathrm{m}$ is mass and $\mathrm{v}$ is speed.

Calculation:

Calculate the wavelength of the bullet by using the formula $\text{λ=}\frac{\text{h}}{\text{mv}}\mathrm{...}\left(\text{1}\right)$

Substitute the given values in equation $\left(1\right)$

  $\begin{array}{c}\mathrm{\lambda }\text{=}\frac{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)}{\left(\text{12}\text{×}{\text{10}}^{\text{-3}}\text{kg}\right)\left(\text{180}\text{m/s}\right)}\\ \text{=}\text{3}\text{.1}\text{×}{\text{10}}^{\text{-34}}\text{m}\end{array}$

(b)

To determine

The minimum uncertainty in its momentum.

Answer to Problem 45GP

  $\text{1}\text{.8}\text{×}{\text{10}}^{\text{-32}}\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given info:

The position of bullet is $\text{0}\text{.60}\text{cm}\text{=}\text{0}\text{.60}{\text{×10}}^{\text{-2}}\text{m}$

Formula used:

  $\mathrm{\Delta }{\mathrm{p}}_{\mathrm{y}}\ge \frac{\mathrm{\hslash }}{\mathrm{\Delta }\mathrm{y}}$

Where, $\mathrm{\hslash }$ is reduced Planck’s constant, $1.055\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$ , $\mathrm{\Delta }{\mathrm{p}}_{\mathrm{y}}$ is uncertainty in momentum and $\mathrm{\Delta }\mathrm{y}$ is the position measurement.

Calculation:

The uncertainty in momentum component perpendicular to motion:

  $\mathrm{\Delta }{\mathrm{p}}_{\mathrm{y}}\ge \frac{\mathrm{h}}{\mathrm{\Delta }\mathrm{y}}\mathrm{...}\left(1\right)$

Substitute the given values in equation $\left(1\right)$

  $\text{=}\frac{\left(\text{1}{\text{.055×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)}{\left(\text{0}\text{.60}{\text{×10}}^{\text{-2}}\text{m}\right)}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-32}}\text{kg}\cdot \text{m/s}$