Chapter 6, Problem 6.15PAE

(a)

Interpretation Introduction

To determine:

Find out the wave length and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is $6.0\times {10}^{-19}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=6.0\times {10}^{-19}J\end{array}$

Putting these values in above equation you get

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{6.0\times {10}^{-19}J}\\ =331.7nm\end{array}$

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

(b)

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is $8.7\times {10}^{-22}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=8.7\times {10}^{-22}J\end{array}$

Putting these values in above equation you get.

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{8.7\times {10}^{-22}J}\\ =2.28758\times {10}^{-4}m\\ =228.7\mu m\end{array}$

With the help of electromagnetic radiation spectrum, this wave length lies in the region of $IR$.

(c)

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is $3.2\times {10}^{-24}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=3.2\times {10}^{-24}J\end{array}$

Putting these values in above equation you get

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m/s\right)}{3.2\times {10}^{-24}J}\\ =0.06219375\\ =62.19mm\end{array}$

With the help of electromagnetic spectrum this wave length lies in the region micro wave.

(d)

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is $1.9\times {10}^{-28}J$.

Explanation of Solution

$\begin{array}{c}\lambda =\left(\frac{hC}{E}\right)\\ h=6.634\times {10}^{-34}Js\\ C=3\times {10}^{8}m/s\\ E=1.9\times {10}^{-28}J\end{array}$

$\begin{array}{c}\lambda =\frac{\left(6.634\times {10}^{-34}Js\right)\times \left(3\times {10}^{8}m.s^{-1}\right)}{1.9\times {10}^{-28}J}\\ =1047.47m\\ =1.047km\end{array}$

With the help of electromagnetic spectrum this wave length lies in the region of long radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.