Chapter 6, Problem 6.17PAE

6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) $3.5\times {10}^{-20}$ J, (b) $8.7\times {10}^{-26}$ J, (c) $7.1\times {10}^{-17}$ J, (d) $5.5\times {10}^{-27}$ J

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is $3.5\times {10}^{-20}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=3.5\times {10}^{-20}J\end{array}$

Putting these values in above equation you get

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{3.5\times {10}^{-20}J}\\ =5.686\times {10}^{-6}m\\ =5.7\mu m\end{array}$

From the electromagnetic spectrum this wave length lies in the $IR$ region.

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is $8.7\times {10}^{-26}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=8.7\times {10}^{-26}J\end{array}$

Putting these values in above equation you get.

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{8.7\times {10}^{-26}J}\\ =2.287\\ =2.3m\end{array}$

From the electromagnetic spectrum this wave length lies in the region of radio wave.

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is $7.1\times {10}^{-17}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=7.1\times {10}^{-17}J\end{array}$

Putting these values in above equation you get

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{7.1\times {10}^{-17}J}\\ =2.803\times {10}^{-9}m\\ =2.803nm\end{array}$

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is $5.5\times {10}^{-17}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=5.5\times {10}^{-27}J\end{array}$

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{5.5\times {10}^{-27}J}\\ =36.18m\end{array}$

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.