Chapter 6, Problem 104AP

Interpretation Introduction

Interpretation:

The de Broglie wavelength (in nm) for the given conditionsis to be determined.

Concept introduction:

The de Broglie wavelength can be determined as follows:

$\lambda =\frac{\text{h}}{mu}$

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

$\text{Conversion factor for second to hour is:}\frac{3600\text{ s}}{1\text{ h}}$

$\text{Conversion factor for nm to meter is}\frac{1.0\text{ nm}}{1.0\times {10}^{-9}\text{ m}}$

$\text{Conversion factor for mile to meter is: }\frac{1609\text{meter}}{1\text{ mile}}$

Answer to Problem 104AP

Solution:

(a)

$1.05\times {10}^{-25}\text{ nm}$

(b)

$\text{8}\text{.6 nm}$

Explanation of Solution

a)Calculate the wavelength of $0.141\text{ kg}$ baseball ((in nm) at this speed

The mass of the ball is $0.141\text{ kg}$.

The speed of the ball is $100\text{ mph }$.

One mile is equivalent to 1609 meters.

Calculate the wavelength (in nm) of the baseball, as shown below:

$\lambda =\frac{\text{h}}{mu}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\text{h}$, $0.141\text{ kg}$ for $m$, and $100\text{ mph}$ for $u$,

$\begin{array}{c}\lambda =\left(\frac{6.63\times {10}^{-34}\cancel{\text{J}}\cdot \cancel{\text{s}}}{0.141\cancel{\text{kg}}\times 100\cancel{\text{mile}}\text{/}\cancel{\text{h}}}\right)\left(\frac{1\cancel{\text{kg}}\cancel{\text{m}}\text{.m }\cancel{{\text{s}}^{-1}}.\cancel{{\text{s}}^{-1}}}{1\cancel{\text{J}}}\right)\left(\frac{1\cancel{\text{mile}}}{1609\cancel{\text{m}}}\right)\left(\frac{1\text{ nm}}{1.0\times {10}^{-9}\cancel{\text{m}}}\right)\left(\frac{3600\cancel{\text{s}}}{1\cancel{\text{h}}}\right)\\ =1.05\times {10}^{-25}\text{ nm}\end{array}$

Therefore, the wavelength is $1.05\times {10}^{-25}\text{ nm}$.

b)Calculate the wavelength (in nm) of the hydrogen atom at the same speed

The mass of the hydrogen atom is $1.007\text{ amu}$.

The speed of the hydrogen atom is $100\text{ mph}$.

One mile is equivalent to $1609\text{ meters}$.

The mass of the hydrogen atom $\left(m\right)$ is $1.007\text{ amu}$.

Calculate the wavelength (in nm) of the hydrogen atom, as shown below:

$\lambda =\frac{\text{h}}{mu}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\text{h}$, $1.007\text{ amu}$ for $m$ and $100\text{ mph}$ for $u$,

$\begin{array}{c}\lambda =\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{\text{1}\text{.007 u}\times 100\text{ mph}}\\ =\left(\frac{6.63\times {10}^{-34}\cancel{\text{J}}\cdot \cancel{\text{s}}}{\text{1}\text{.007 }\cancel{\text{u}}\times 100\cancel{\text{mile}}\text{/}\cancel{\text{h}}}\right)\left(\frac{1\cancel{\text{u}}}{1.66\times {10}^{-27}\cancel{\text{kg}}}\right)\left(\frac{1\cancel{\text{kg}}\cancel{\text{m}}\text{.}\cancel{\text{m}}\cancel{{\text{s}}^{-1}}\text{.}\cancel{{\text{s}}^{-1}}}{1\cancel{\text{J}}}\right)\left(\frac{1\cancel{\text{mile}}}{1609\cancel{\text{m}}}\right)\left(\frac{1\text{ nm}}{1.0\times {10}^{-9}\cancel{\text{m}}}\right)\left(\frac{3600\cancel{\text{s}}}{1\cancel{\text{h}}}\right)\\ =8.86\text{ nm}\end{array}$

Therefore, the wavelength is $8.86\text{ nm}$.