Chapter 7, Problem 7A.3E

(a)

Interpretation Introduction

Interpretation:

The quantum size involved in excitation with frequency of $1\times {10}^{15}Hz$ has to be calculated and expressed in both joules and in $\text{kJ/mol}$ units.

Concept Introduction:

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

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

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules.

Explanation of Solution

Given:

Frequency is $1\times {10}^{15}Hz$.

The quanta transited in an electronic transition of known frequency is equal to the respective photon energy.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules.

$\begin{array}{c}{\text{E}}_{\text{photon}}\text{=}\text{hν}\\ \text{=}\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.s×}\left({\text{1×10}}^{\text{15}}{\text{s}}^{\text{-1}}\right)\\ =6.6\times {10}^{-19}\text{J}\end{array}$

The value is obtained in $\text{kJ/mol}$ by multiplying the photon energy with Avogadro’s number.

$\begin{array}{c}{\text{E}}_{\text{m}}\text{=}{\text{N}}_{\text{A}}{\text{E}}_{\text{photon}}\\ \text{=}\text{6}{\text{.022×10}}^{23}{\text{mol}}^{\text{-1}}\text{×}\left(\text{6}{\text{.6×10}}^{\text{-19}}\text{J}\right)\\ \text{=}397452\text{J}{\text{mol}}^{\text{-1}}\\ =4{\text{×10}}^2\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The quantum size involved in molecular vibration with period of $20fs$ has to be calculated.

Concept Introduction:

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

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

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules.

Explanation of Solution

Given:

Period is $20fs$.

The quanta transited in an electronic transition of known frequency is equal to the respective photon energy.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules. The time period is equal to inverse of frequency.

$\begin{array}{c}\Delta \text{E}\text{=}\text{h}\times \frac{1}{\text{T}}\text{since,}\text{T=}\frac{\text{1}}{\text{ν}}\\ \text{=}\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.s×}\left({\text{20×10}}^{\text{-15}}\text{s}\right)\left[here,{\text{1fs = 10}}^{\text{-15}}\text{s}\right]\\ =3.3\times {10}^{-20}\text{J}\end{array}$

The value is obtained in $\text{kJ/mol}$ by multiplying the obtained energy with Avogadro’s number.

$\begin{array}{c}\Delta {\text{E}}_{\text{m}}\text{=}{\text{N}}_{\text{A}}\text{E}\\ \text{=}\text{6}{\text{.022×10}}^{23}{\text{mol}}^{\text{-1}}\text{×}\left(\text{3}{\text{.3×10}}^{\text{-20}}\text{J}\right)\\ \text{=}19872.6\text{J}{\text{mol}}^{\text{-1}}\\ =20\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The quantum size involved in pendulum with period of $0.50s$ has to be calculated.

Concept Introduction:

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

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

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules.

Explanation of Solution

Given:

Period is $0.50s$.

The quanta transited in an electronic transition of known frequency is equal to the respective photon energy.

The photon energy is obtained by multiplication of Planck’s constant and frequency in which energy is obtained in Joules. The time period is equal to inverse of frequency.

$\begin{array}{c}\Delta \text{E}\text{=}\text{h}\times \frac{1}{\text{T}}\text{since,}\text{T=}\frac{\text{1}}{\text{ν}}\\ \text{=}\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{.s×}\left(0.5\text{s}\right)\\ =1.3\times {10}^{-33}\text{J}\end{array}$

The value is obtained in $\text{kJ/mol}$ by multiplying the obtained energy with Avogadro’s number.

$\begin{array}{c}\Delta {\text{E}}_{\text{m}}\text{=}{\text{N}}_{\text{A}}\text{E}\\ \text{=}\text{6}{\text{.022×10}}^{23}{\text{mol}}^{\text{-1}}\text{×}\left(\text{1}{\text{.3×10}}^{\text{-33}}\text{J}\right)\\ =7.8\times {10}^{-13}\text{kJ}{\text{mol}}^{\text{-1}}here,1\text{kJ}={10}^3\text{J}\end{array}$