Chapter 3, Problem 3.14QA

Interpretation Introduction

To explain:

with a sketch why the frequencies of long-wavelength waves of electromagnetic radiations are lower than those of short-wavelength waves.

Answer to Problem 3.14QA

Solution:

The frequencies of long-wavelength waves of electromagnetic radiations are lower than those of short-wavelength waves.

Explanation of Solution

1) Formula:

i. $\mathrm{\nu }=\frac{\mathrm{C}}{\mathrm{\lambda }}$

ii. $\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }[\mathrm{ }\frac{1}{{\mathrm{n}\mathrm{f}}^2}-\frac{1}{{\mathrm{n}\mathrm{i}}^2}]$

2) Concept:

Frequency, wavelength and energy are related to each other by the above formula.

Let’s take an example of transition of electron from (a) n=3 to n=2 and (b) n= 3 to n=1

2. Calculations:

$\mathrm{ }\left(\mathrm{a}\right)\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }[\mathrm{ }\frac{1}{2^2}-\frac{1}{3^2}]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }[\mathrm{ }0.25-0.1111]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }\left[\mathrm{ }0.1389\right]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.52\times {10}^6{\mathrm{m}}^{-1}$

$\mathrm{\lambda }=\frac{1}{1.52\times {10}^6{\mathrm{m}}^{-1}}$

$\mathrm{\lambda }=6.58\times {10}^{-7\mathrm{ }}\mathrm{m}$

$\mathrm{\nu }=\frac{2.998\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}}{6.58\times {10}^{-7\mathrm{ }}\mathrm{m}}$

$\mathrm{\nu }=4.56\times {10}^{14}\mathrm{H}\mathrm{z}$

$\left(\mathrm{b}\right)\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }[\mathrm{ }\frac{1}{1^2}-\frac{1}{3^2}]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }[\mathrm{ }1-0.1111]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }1.097\times \mathrm{ }{10}^7\mathrm{ }{\mathrm{m}}^{-1}\mathrm{ }\left[0.8889\right]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{ }9.75\times {10}^6{\mathrm{m}}^{-1}$

$\mathrm{\lambda }=\frac{1}{9.75\times {10}^6{\mathrm{m}}^{-1}}$

$\mathrm{\lambda }=1.03\times {10}^{-7\mathrm{ }}\mathrm{m}$

$\mathrm{\nu }=\frac{2.998\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}}{1.03\times {10}^{-7\mathrm{ }}\mathrm{m}}$

$\mathrm{\nu }=2.91\times {10}^{15}\mathrm{H}\mathrm{z}$

From the sketch and calculations, we can say that shorter transition of electron (n=3 to n=2), the wavelength is large thus frequency is short. And for longer transition of electron (n=3 to n=1), the wavelength is small therefore the frequency is large.

Conclusion:

The frequencies of long-wavelength wave of electromagnetic radiations are lower than those of short-wavelength waves.