Chapter 27.1, Problem 3PP

To determine

To Rank: Photon in increasing order on the basis of energy.

Answer to Problem 3PP

  $\text{C}\text{<}\text{D}\text{<}\text{B}\text{<}\text{A}$

Explanation of Solution

Given data:

  $\begin{array}{l}{E}_A=4.0\text{ eV}\\ {\lambda }_B\text{= }320\text{ nm}\\ {\lambda }_C\text{= }811\text{ nm}\\ E_D\text{ = }2.1\text{ eV}\end{array}$

There are four photons given here. Energy of two photons are given and remaining two photons’ wavelengths are given. Find the energy of the photon whose wavelengths are given.

Formula used:

Energy of the photon can be obtained by:

  $E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }$

Where, $\lambda$ is the wavelength.

Calculation:

Energy of photon B can be calculated as follows:

  $\begin{array}{l}{E}_B=\frac{1240\text{eV}\cdot \text{nm}}{{\lambda }_B}\\ E_B=\frac{1240\text{eV}\cdot \text{nm}}{320\text{nm}}=3.875\text{eV}\end{array}$

Energy of photon C can be calculated as follows:

  $\begin{array}{l}{E}_C=\frac{1240\text{eV}\cdot \text{nm}}{{\lambda }_C}\\ E_C=\frac{1240\text{eV}\cdot \text{nm}}{811\text{nm}}=1.53\text{eV}\end{array}$

Conclusion:

The rank of photons in ascending order of energy is $\text{C}\text{<}\text{D}\text{<}\text{B}\text{<}\text{A}$ .