Chapter 12, Problem 125E

Interpretation Introduction

Interpretation:Frequency and energy of photon for two different wavelengths is to be calculated.

Concept introduction:Energy possessed by a single photon is termed as photon energy. Energy of photon is calculated as follows:

  $E_{\text{photon}}=h\nu =\frac{hc}{\lambda }$

Where,

  $E_{\text{photon}}$ is energy of photon.

  $h$ isPlanck’s constant.

  $\nu$ is frequency of photon.

  $c$ is speed of light.

  $\lambda$ is wavelength of energy.

Energy of photon is directly related to frequency but inversely related to wavelength of light.

Answer to Problem 125E

For wavelength $589\text{ nm}$ , frequency of photon is $5.090\times {10}^{14}{\text{ s}}^{-1}$ and energy is $3.373\times {10}^{-19}\text{ J}$ and $203.1\text{ kJ/mol}$ .

For wavelength $589.6\text{ nm}$ , frequency of photon is $5.085\times {10}^{14}{\text{ s}}^{-1}$ and energy is $3.369\times {10}^{-19}\text{ J}$ and $202.9\text{ kJ/mol}$ .

Explanation of Solution

Given Information:Value of wavelength for one emission line is $589\text{ nm}$ and second emission line is $589.6\text{ nm}$ .

Frequency of photon of wavelength $589\text{ nm}$ is calculated as follows:

  $\nu =\frac{c}{\lambda }$

Where,

• $\nu$ is frequency of photon.
• $c$ is speed of light.
• $\lambda$ is wavelength of energy.

Value of $c$ is $3\times {10}^8\text{ m/s}$ .

Value of $\lambda$ is $589\text{ nm}$ .

Substitute values in above equation.

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{\left(3\times {10}^8\text{ m/s}\right)}{\left(589\text{ nm}\right)}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =5.090\times {10}^{14}{\text{ s}}^{-1}\end{array}$

Thus, frequency of photon is $5.090\times {10}^{14}{\text{ s}}^{-1}$ .

Formula to calculate energy is as follows:

  $E_{\text{photon}}=h\nu$

Where,

• $E_{\text{photon}}$ is energy of photon.
• $h$ isplanck’s constant.
• $\nu$ is frequency of photon.

Value of $h$ is $6.626\times {10}^{-34}\text{ Js}$ .

Value of $\nu$ is $5.090\times {10}^{14}{\text{ s}}^{-1}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{\text{photon}}=h\nu \\ =\left(6.626\times {10}^{-34}\text{ Js}\right)\left(5.090\times {10}^{14}{\text{ s}}^{-1}\right)\\ =3.373\times {10}^{-19}\text{ J}\end{array}$

Hence, energy of photon is $3.373\times {10}^{-19}\text{ J}$ .

Energy in $\text{kJ/mol}$ is calculated as follows:

  $E\left(\text{kJ/mol}\right)=E\left(\text{J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)$

Where,

• $E\left(\text{J}\right)$ is energy of photon in joule.
• $E\left(\text{kJ/mol}\right)$ is energy of photon in $\text{kJ}$ .

Value of $E\left(\text{J}\right)$ is $3.373\times {10}^{-19}\text{ J}$ .

Substitute valuein above equation.

  $\begin{array}{c}E\left(\text{kJ/mol}\right)=E\left(\text{J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\\ =\left(3.373\times {10}^{-19}\text{ J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\\ =203.1\text{ kJ/mol}\end{array}$

Hence, energy of photon in $\text{kJ/mol}$ is $203.1\text{ kJ/mol}$ .

Frequency of photon of wavelength $589.6\text{ nm}$ is calculated as follows:

  $\nu =\frac{c}{\lambda }$

Value of $c$ is $3\times {10}^8\text{ m/s}$ .

Value of $\lambda$ is $589.6\text{ nm}$ .

Substitute values in above equation.

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{\left(3\times {10}^8\text{ m/s}\right)}{\left(589.6\text{ nm}\right)}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =5.085\times {10}^{14}{\text{ s}}^{-1}\end{array}$

Thus, frequency of photon is $5.085\times {10}^{14}{\text{ s}}^{-1}$ .

Formula to calculate energy is as follows:

  $E_{\text{photon}}=h\nu$

Value of $h$ is $6.626\times {10}^{-34}\text{ Js}$ .

Value of $\nu$ is $5.085\times {10}^{14}{\text{ s}}^{-1}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{\text{photon}}=h\nu \\ =\left(6.626\times {10}^{-34}\text{ Js}\right)\left(5.085\times {10}^{14}{\text{ s}}^{-1}\right)\\ =3.369\times {10}^{-19}\text{ J}\end{array}$

Hence, energy of photon is $3.369\times {10}^{-19}\text{ J}$ .

Energy in $\text{kJ/mol}$ is calculated as follows:

  $E\left(\text{kJ/mol}\right)=E\left(\text{J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)$

Value of $E\left(\text{J}\right)$ is $3.369\times {10}^{-19}\text{ J}$ .

Substitute valuein above equation.

  $\begin{array}{c}E\left(\text{kJ/mol}\right)=E\left(\text{J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\\ =\left(3.369\times {10}^{-19}\text{ J}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\\ =202.9\text{ kJ/mol}\end{array}$

Hence, energy of photon in $\text{kJ/mol}$ is $202.9\text{ kJ/mol}$ .

Conclusion

For wavelength $589\text{ nm}$ , frequency of photon is $5.090\times {10}^{14}{\text{ s}}^{-1}$ and energy is $3.373\times {10}^{-19}\text{ J}$ and $203.1\text{ kJ/mol}$ .

For wavelength $589.6\text{ nm}$ , frequency of photon is $5.085\times {10}^{14}{\text{ s}}^{-1}$ and energy is $3.369\times {10}^{-19}\text{ J}$ and $202.9\text{ kJ/mol}$ .