Chapter 27.1, Problem 4PP

To determine

To Find: Range of energies associated with the photons from the given visible light spectrum.

Answer to Problem 4PP

  $1.77\text{eV to }3.1\text{eV}$

Explanation of Solution

Given data:

From the above given figure, the minimum and maximum wavelengths of the visible spectrum are (range) $400\text{nm}\text{to 700 nm}$ .

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 the visible light at 400 nm is:

  $\begin{array}{l}E=\frac{1240\text{eV}\cdot \text{nm}}{400\text{nm}}\\ E=3.1\text{eV}\end{array}$

Energy of the visible light at 700 nm is:

  $\begin{array}{l}E=\frac{1240\text{eV}\cdot \text{nm}}{400\text{nm}}\\ E=1.77\text{eV}\end{array}$

Conclusion:

The range of energies associated with the photons in the visible light spectrum is $1.77\text{eV to }3.1\text{eV}$ .