Chapter 5, Problem 52A

Interpretation Introduction

Interpretation:

The frequency of photon and type of electromagnetic radiation needs to be determined.

Concept introduction:

Frequency of photon is inversely proportional to the wavelength of photon. And it is directly proportional to the energy of photon.

$E_{photon}=hv$

Answer to Problem 52A

Frequency of photon is $\left(v\right)=4.42\times {10}^8{s}^{-1}$

Type of radiation is Radio waves

Given information:

Energy of photon $=2.93\times {10}^{-25}\text{ J}$

Calculations:

As per Planck- Einstein equation

$E_{photon}=hv$

Where is $E$ is the energy of photon, $h$ refers to Planck’s constant and $v$ refers to frequency

Now putting value of $h=6.626\times {10}^{-34}Js$ And $E_{photon}=2.93\times {10}^{-25}J$

$\begin{array}{l}2.93\times {10}^{-25}J= 6.626\times {10}^{-34}Js\times v\\ v=\frac{2.93\times {10}^{-25}J}{6.626\times {10}^{-34}Js}\\ v=\text{ 0}\text{.442}\times {\text{10}}^9{s}^{-1}\\ v=4.42\times {10}^8{s}^{-1}\end{array}$

This frequency range is for TV or FM that is radio waves.

Thus, $v=4.42\times {10}^8{s}^{-1}$, which is type of radio wave.

Explanation of Solution

Given information:

Energy of photon $=2.93\times {10}^{-25}\text{ J}$

Calculations:

As per Planck- Einstein equation

$E_{photon}=hv$

Where is $E$ is the energy of photon, $h$ refers to Planck’s constant and $v$ refers to frequency

Now putting value of $h=6.626\times {10}^{-34}Js$ And $E_{photon}=2.93\times {10}^{-25}J$

$\begin{array}{l}2.93\times {10}^{-25}J= 6.626\times {10}^{-34}Js\times v\\ v=\frac{2.93\times {10}^{-25}J}{6.626\times {10}^{-34}Js}\\ v=\text{ 0}\text{.442}\times {\text{10}}^9{s}^{-1}\\ v=4.42\times {10}^8{s}^{-1}\end{array}$

This frequency range is for TV or FM that is radio waves.

Conclusion

Thus, $v=4.42\times {10}^8{s}^{-1}$, which is type of radio wave.