Chapter 5.1, Problem 7PP

Interpretation Introduction

Interpretation:

Energy of one photon of the microwave radiation needs to be calculated.

Concept introduction:

The energy of photon is directly proportional to frequency. This is represented as follows:

$E\propto \nu$ or $E=hv=\frac{hc}{\lambda }$

Where, $h=$ Planck's constant $=6.62\times {10}^{-27}erg.\sec$ or $6.62\times {10}^{-34}Joules\sec .$

Answer to Problem 7PP

The energy of one photon is $1.59\times {10}^{-24}Joule$

Explanation of Solution

Given information:

Wavelength $=0.125m$

Calculation:

$E=\frac{hc}{\lambda }$

Where,

E = Energy of photon

$h=$ Planck's constant $=6.62\times {10}^{-27}erg.\sec$ or $6.62\times {10}^{-34}Joules\sec .$

c = velocity of light $\left(3\times {10}^{8}m/s\right)$

$\begin{array}{l}E=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{125\times {10}^{-3}m}\\ E=\frac{6.626\times 3\times {10}^{-34+8+3}}{125}\\ E=0.159\times {10}^{-23}Joule\\ E=1.59\times {10}^{-24}Joule\end{array}$

Conclusion

The energy of one photon is $1.59\times {10}^{-24}Joule$