Chapter 5.1, Problem 6PP

Interpretation Introduction

Interpretation:

Energy of one photon 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 6PP

The energy of one photon = $4.417\times {10}^{-19}\mathrm{Joule}$

Explanation of Solution

Given information:

Wavelength $=4.50\times {10}^{2}nm=4.50\times {10}^2\times {10}^{-9}m$

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

$\lambda$ = wavelength

Putting the values,

$\begin{array}{l}=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{4.50\times {10}^2\times {10}^{-9}m}\\ =\frac{6.626\times 3\times {10}^{-34+8+7}}{4.5}\\ =\frac{6.626\times 3\times {10}^{-19}}{4.5}\\ =4.417\times {10}^{-19}\mathrm{Joule}\end{array}$

Conclusion

The energy of one photon = $4.417\times {10}^{-19}\mathrm{Joule}$