Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy $\left(\text{J/photon and kJ/mol}\right)$ of photon with $\text{250 nm}$ have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

  $E=\frac{hc}{\lambda }$

Here,

$\lambda$ denotes wavelength.

$h$ denotes Planck’s constant

$c$ denotes speed of light.

Explanation of Solution

Conversion factor to convert $\text{nm}$ to $\text{m}$ is as follows:

  $1\text{ nm}={10}^{-9}\text{ m}$

Thus, $\text{250 nm}$ is converted to $\text{m}$ as follows:

  $\begin{array}{c}\text{Wavelength}=\left(\text{250 nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\\ =2.5\times {10}^{-7}\text{ m}\end{array}$

Formula to calculate frequency is as follows:

  $v=\frac{c}{\lambda }$        (1)

Substitute $2.5\times {10}^{-7}\text{ m}$ for $\lambda$ , and $3\times {10}^8\text{ m/s}$ for $c$ in equation (1).

  $\begin{array}{c}v=\frac{3\times {10}^8\text{ m/s}}{2.5\times {10}^{-7}\text{ m}}\\ =1.2\times {10}^{15}\text{ Hz}\end{array}$

Formula to calculate wavenumber is as follows:

  $\overline{v}=\frac{v}{c}$        (2)

Substitute $1.2\times {10}^{15}\text{ Hz}$ for $v$ , and $3\times {10}^8\text{ m/s}$ for $c$ in equation (2).

  $\begin{array}{c}\overline{v}=\left(\frac{1.2\times {10}^{15}\text{ Hz}}{3\times {10}^8\text{ m/s}}\right)\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\\ =4.0\times {10}^4{\text{ cm}}^{-1}\end{array}$

Planck’s hypothesis states energy associated with single photon is as follows:

  $E=\frac{hc}{\lambda }$        (3)

Substitute $2.5\times {10}^{-7}\text{ m}$ for $\lambda$ , $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (3).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\left(2.5\times {10}^{-7}\text{ m}\right)}\\ =7.95\times {10}^{-19}\text{ J/photon}\end{array}$

Avogadro’s number is $6.022\times {10}^{23}\text{ photons/mol}$. So energy for $1\text{ mol}$ photons is calculated as follows:

  $\begin{array}{c}E=\left(6.022\times {10}^{23}\text{ photons/mol}\right)\left(\frac{7.95\times {10}^{-19}\text{ J}}{1\text{ photon}}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ =478.7\text{ kJ/mol}\\ \approx 479\text{ kJ/mol}\end{array}$

Thus, frequency, wavenumber and energy $\left(\text{J/photon and kJ/mol}\right)$ are $1.2\times {10}^{15}\text{ Hz}$, $4.0\times {10}^3{\text{ cm}}^{-1}$, $7.95\times {10}^{-19}\text{ J/photon}$ and $479\text{ kJ/mol}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy $\left(\text{J/photon and kJ/mol}\right)$ of photon with $\text{2}\text{.50 μm}$ have to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Conversion factor to convert $\text{μm}$ to $\text{m}$ is as follows:

  $1\text{ μm}={10}^{-6}\text{ m}$

Thus, $\text{2}\text{.50 μm}$ is converted to $\text{m}$ as follows:

  $\begin{array}{c}\text{Wavelength}=\left(\text{2}\text{.50 μm}\right)\left(\frac{{10}^{-6}\text{ m}}{1\text{ μm}}\right)\\ =2.5\times {10}^{-6}\text{ m}\end{array}$

Substitute $2.5\times {10}^{-6}\text{ m}$ for $\lambda$ , and $3\times {10}^8\text{ m/s}$ for $c$ in equation (1).

  $\begin{array}{c}v=\frac{3\times {10}^8\text{ m/s}}{2.5\times {10}^{-6}\text{ m}}\\ =1.2\times {10}^{14}\text{ Hz}\end{array}$

Substitute $1.2\times {10}^{14}\text{ Hz}$ for $v$ , and $3\times {10}^8\text{ m/s}$ for $c$ in equation (2).

  $\begin{array}{c}\overline{v}=\left(\frac{1.2\times {10}^{14}\text{ Hz}}{3\times {10}^8\text{ m/s}}\right)\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\\ =4.0\times {10}^3{\text{ cm}}^{-1}\end{array}$

Substitute $2.5\times {10}^{-6}\text{ m}$ for $\lambda$ , $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (3).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\left(2.5\times {10}^{-6}\text{ m}\right)}\\ =7.95\times {10}^{-20}\text{ J/photon}\end{array}$

Avogadro’s number is $6.022\times {10}^{23}\text{ photons/mol}$. So energy for $1\text{ mol}$ photons is calculated as follows:

  $\begin{array}{c}E=\left(6.022\times {10}^{23}\text{ photons/mol}\right)\left(\frac{7.95\times {10}^{-20}\text{ J}}{1\text{ photon}}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ =47.87\text{ kJ/mol}\\ \approx \text{47}\text{.9 kJ/mol}\end{array}$

Thus, frequency, wavenumber and energy $\left(\text{J/photon and kJ/mol}\right)$ are $1.2\times {10}^{14}\text{ Hz}$, $4.0\times {10}^3{\text{ cm}}^{-1}$, $7.95\times {10}^{-20}\text{ J/photon}$ and $\text{47}\text{.9 kJ/mol}$ respectively.