Chapter 7, Problem 7A.5E

(a)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for given mass of one gram and speed of $\text{1}{\text{ms}}^{\text{-1}}$ has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

Explanation of Solution

Given:

Mass $\text{=}\text{1}\text{g}$

Speed $\text{=}\text{1}{\text{ms}}^{\text{-1}}$

The de Broglie wavelength is calculated by using de Broglie relation.

$\begin{array}{c}\text{ λ}\text{=}\frac{\text{h}}{\text{p}}\\ \text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\left[here,\text{p}=\text{m}\text{υ}\right]\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{1\times {10}^{-3}\text{kg}\times \text{1}{\text{ms}}^{\text{-1}}}\left[here,1\text{g}={10}^{-3}\text{kg}\right]\\ =6.626\times 10^{-31}m\\ \end{array}$

(b)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for given mass of one gram and speed of $\text{1}\times {10}^5\text{km}{\text{s}}^{\text{-1}}$ has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

Explanation of Solution

Given:

Mass $\text{=}\text{1}\text{g}$

Speed $\text{=}\text{1}\times {\text{10}}^{\text{5}}k\text{m}{\text{s}}^{\text{-1}}$

The de Broglie wavelength is calculated by using de Broglie relation.

$\begin{array}{c}\text{ λ}\text{=}\frac{\text{h}}{\text{p}}\\ \text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\left[here,\text{p}=\text{m}\text{υ}\right]\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{1\text{g}\times \text{1}\times {10}^5{\text{kms}}^{\text{-1}}}\left[\begin{array}{l}here,1\text{g}={10}^{-3}\text{kg}\\ \text{1km}={10}^3\text{m}\end{array}\right]\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{1\times {10}^{-3}\text{kg}\times \text{1}\times {10}^8{\text{ms}}^{\text{-1}}}\\ =6.63\times {10}^{-39}\text{m}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for helium atom with speed of $\text{1}000{\text{ms}}^{\text{-1}}$ has to be calculated.

Concept Introduction:

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\end{array}$

h is Planck’s constant($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$) which relates energy and frequency.

$\text{υ}$ is the speed of particle.

m is the mass of particle.

$\text{λ}$ is the wavelength.

p is the momentum.

The above equation is called de Broglie relation.

Explanation of Solution

Given:

Mass $\text{He}\text{=}\text{4}\text{u}$

Speed $\text{=}\text{1000}{\text{ms}}^{\text{-1}}$

The de Broglie wavelength is calculated by using de Broglie relation.

$\begin{array}{c}\text{ λ}\text{=}\frac{\text{h}}{\text{p}}\\ \text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\left[here,\text{p}=\text{m}\text{υ}\right]\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{4\text{u}\times \text{1}000{\text{ms}}^{\text{-1}}}\left[here,1\text{u}=1.6605\times {10}^{-27}\text{kg}\right]\\ =\frac{6.626\times {10}^{-34}\text{J}\text{.s}}{4\times 1.6605\times {10}^{-27}\text{kg}\times \text{1}000{\text{ms}}^{\text{-1}}}\\ =\frac{6.626\times {10}^{-34}}{6.642\times {10}^{-24}}\\ =9.97\times {10}^{-11}\text{m}\\ \text{=}99.7\text{pm}\left[here,1\text{pm}={10}^{-12}\text{m}\right]\end{array}$