Chapter 5, Problem 57A

Interpretation Introduction

Interpretation:

The frequency of the $ArF$ laser’s radiation and energy of a single quantum of the radiation needs to be determined.

Concept introduction:

As per Planck Einstein equation

$E_{photon}=hv$

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

Answer to Problem 57A

Frequency of the $ArF$ laser’s radiation $\left(v\right)=1.55\times {10}^{15}Hz$

$E_{photon}=1.03\times {10}^{-18}J$

Explanation of Solution

Given:

Wavelength of $ArF$ ’s electromagnetic radiation $=193.3nm$.

Explanations:

As per wave equation of motion:

$c=\lambda \times v$

Here, c represents velocity and $v$ represents the frequency.

Now putting the value of $\lambda =1.933\times {10}^{-7}m$ and $c=3\times {10}^{8}m{s}^{-1}$

The frequency can be calculated as follows:

$\frac{c}{\lambda }=v$

Putting the values,

$v=\frac{3\times {10}^8\text{ m/s}}{1.933\times {10}^{-7}\text{ m}}\text{ =}1.55\times {10}^{15}{s}^{-1}$

Now, as per Planck Einstein equation

$E_{photon}=hv$

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

Now putting the value of $h=6.626\times {10}^{-34}Js$ and $v=1.55\times {10}^{15}{\text{Hz or s}}^{-1}$

$E_{photon}=\left(6.626\times {10}^{-34}\text{ Js}\right)\left(1.55\times {10}^{15}{\text{ s}}^{-1}\right)=1.03\times {10}^{-18}\text{ J}$

Conclusion

Frequency of the $ArF$ laser’s radiation $\left(v\right)=1.55\times {10}^{15}Hz$

$E_{photon}=1.03\times {10}^{-18}J$