Chapter 5, Problem 5.43SP

Interpretation Introduction

To determine:

The energy of each of the given photons in kilojoules per mole.

Solution:

1. $\text{E = 2}{\text{.38 ×10}}^{\text{7}}\frac{\text{kJ}}{\text{mol}}$
2. $\text{E = 5}\text{.03 }\times {\text{10}}^{\text{-7}}\frac{\text{kJ}}{\text{mol}}$
3. $\text{E = 4}\text{.66 }\times {\text{10}}^{\text{-7}}\frac{\text{kJ}}{\text{mol}}$

Explanation:

Given:

  $\text{v = 5}\text{.97 }\times {\text{10}}^{\text{19}}{\text{ s}}^{\text{-1}}$

  $\text{v = 1}\text{.26 }\times {\text{10}}^{\text{6}}{\text{ s}}^{\text{-1}}$

  $\text{λ =2}{\text{.57 x 10}}^{\text{2}}\text{ m}$

The energy emitted or absorbed is quantized. This means that the energy is emitted in packages that have the following form,

   $\text{E = hv}$

where $\text{h}$ is the Planck’s constant and $\text{v}$ is the frequency.
The energy emitted or absorbed is a multiple of $hv$. So that, $hv$ is multiplied by a whole number $n$.
When n is equal to the Avogadro’s number ( ${\text{N}}_{\text{A}}$ ), one mole of photons emits or absorbs a determined energy per mol.

Formula used:

  ${\text{E = N}}_{\text{A}}\text{hv}$

where $\text{E}$ is the energy emitted or absorbed, ${\text{N}}_{\text{A}}$ is the Avogadro’s number ( $\text{6}\text{.02}\times {\text{10}}^{\text{23}}\text{ particles/mol}$ ), $\text{h}$ is the Planck’s constant ( $\text{ 6}\text{.626 }\times {\text{ 10}}^{\text{-37}}\text{ kJ s}$ ) and $\text{v}$ is the frequency.

  $\text{λ = }\frac{\text{c}}{\text{v}}\to \text{v = }\frac{\text{c}}{\text{λ}}$

where $\text{λ}$ is the wavelength, $\text{c}$ is the speed of light ( ${\text{3x10}}^{\text{8}}\text{ m/s}$ ) and $\text{v}$ is the frequency.

So that,

   ${\text{E = N}}_{\text{A}}\text{h}\frac{\text{c}}{\text{λ}}$

Calculation:

1. $\text{E = 6}\text{.02 }\times {\text{10}}^{\text{23}}\frac{\text{photon}}{\text{mol}}\times \text{ 6}\text{.626 }\times {\text{10}}^{\text{-37}}\text{ kJ s }\times \text{ 5}\text{.97}\times {\text{ 10}}^{\text{19}}{\text{ s}}^{\text{-1}}$

   $\text{E = 2}{\text{.38 × 10}}^{\text{7}}\frac{\text{kJ}}{\text{mol}}$

2. $\text{E = 6}\text{.02 }\times {\text{ 0}}^{\text{23}}\frac{\text{photon}}{\text{mol}}\times \text{ 6}{\text{.626 x 10}}^{\text{-37}}\text{ kJ s }\times \text{ 1}\text{.26 }\times {\text{ 10}}^{\text{6}}{\text{ s}}^{\text{-1}}$

   $\text{E = 5}{\text{.03 x 10}}^{\text{-7}}\frac{\text{kJ}}{\text{mol}}$

3. $\text{E = 6}\text{.02 }\times {\text{10}}^{\text{23}}\frac{\text{photon}}{\text{mol}}\times \text{ 6}\text{.626 }\times {\text{10}}^{\text{-37}}\text{ kJ s }\times \frac{{\text{3 x 10}}^{\text{8}}\frac{\text{m}}{\text{s}}}{\text{2}{\text{.57 x 10}}^{\text{2}}\text{ m}}$

   $\text{E = 4}{\text{.66 × 10}}^{\text{-7}}\frac{\text{kJ}}{\text{mol}}$

Conclusion:

1. The photon has an energy of $2.38\text{ x }{10}^7\frac{\text{kJ}}{\text{mol}}$
2. The photon has an energy of $5.03\text{ x }{10}^{-7}\frac{\text{kJ}}{\text{mol}}$
3. The photon has an energy of $4.66\text{ x }{10}^{-7}\frac{\text{kJ}}{\text{mol}}$