Question

Radiation leaves a source at a natural frequency of (a) 9.10 x 10¹³Hz. What is the wavenumber corresponding to this frequency? If a Michelson interferometer has a mirror velocity of 1.00 cm/s, what will the frequency be at the transducer. Find the wavenumber at the source and frequency at the transducer for source radiation of

(b) 9.20 x 10¹³Hz and

(c) 9.30 x 10¹³ Hz.

Short answer

a) Wavenumber at the given frequency is 3.03 x 10³cm-1 and frequency of the transducer is 6.07 kHz.
b) Wavenumber at the given frequency is 3.07x 10³cm-1 and frequency of the transducer is 6.13 kHz

c) Wavenumber at the given frequency is 3.10x 10³cm-1 and frequency of the transducer is 6.20 kHz

Step-by-step solution

Concept

Frequency number of the occurrences of a particular event in a unit time. Wave number is a unit of frequency, often used in atomic, molecular, and nuclear spectroscopy.

The absorption maximum is given as :
$\overline{v}=\frac{1}{2\pi c}\sqrt{\frac{k}{\mu }}$
Where c is the velocity of light

The frequency is given by

$f=2v_{\mathrm{M}}\overline{v}\dots \dots \text{(II)}$
Where the
f=frequency
v_M=the constant and
$\overline{v}$=wavenumber

Substitute role="math" localid="1645888139139" $v=\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}$ in Equation (I), we get\

$\overline{v}=\frac{v}{c}\dots \dots \text{(III)}$
Substitute for role="math" localid="1645888210654" $\overline{v}=\frac{v}{c}$in Equation (II), we get
$f=\frac{2v_{\mathrm{M}}v}{c}\text{......(IV)}$

Part(a) Step 1- Given 

Given natural frequency is 9.1x10¹³ Hz.

Mirror velocity is 1 cm/sec.

Find the wavenumber at the source and frequency at the transducer

2- Part(a) Step2-Explanation.

Substitute9.1x10¹³ Hz and c=10¹⁰ cm/sec in equation (III),

$$\begin{gathered} \overline{V}=\frac{9.10\times {10}^{13}\mathrm{Hz}}{3\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =\frac{9.10}{3}\times {10}^3{\mathrm{cm}}^{-1} \\ =3.03\times {10}^3{\mathrm{cm}}^{-1} \end{gathered}$$

Use this wave number and v_M = 1 cm/sec in the equation (IV),

$$\begin{gathered} f=\frac{2\left(1.00{\mathrm{cms}}^{-1}\right)\left(9.10\times {10}^{13}\mathrm{Hz}\right)}{3.00\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =6.07\times {10}^3{\mathrm{s}}^{-1}\left(\frac{1\mathrm{kHz}}{{10}^3{\mathrm{s}}^{-1}}\right) \\ =6.07\mathrm{kHz} \end{gathered}$$

Part(b) Step 1- Given 

Given natural frequency is 9.20 x 10¹³ Hz.

Mirror velocity is 1 cm/sec.

Find the wavenumber at the source and frequency at the transducer

Part(b) Step 2 - Explanation

Substitute 9.20 x 10¹³ Hz and c=10¹⁰ cm/sec in equation (III),

$$\begin{gathered} \overline{v}=\frac{9.20\times {10}^{13}\mathrm{Hz}}{3\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =\frac{9.20}{3}\times {10}^3{\mathrm{cm}}^{-1} \\ =3.07\times {10}^3{\mathrm{cm}}^{-1} \end{gathered}$$

Use this wave number and v_M = 1 cm/sec in the equation (IV),

$$\begin{gathered} f=\frac{2\left(1.00{\mathrm{cms}}^{-1}\right)\left(9.20\times {10}^{13}\mathrm{Hz}\right)}{3.00\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =6.13\times {10}^3{\mathrm{s}}^{-1}\left(\frac{1\mathrm{kHz}}{{10}^3{\mathrm{s}}^{-1}}\right) \\ =6.13\mathrm{kHz} \end{gathered}$$

Part(c) Step1 Given: 

Given natural frequency is 9.30 x 10¹³ Hz.

Mirror velocity is 1 cm/sec.

Find the wavenumber at the source and frequency at the transducer

Part(c) Step2 -Explanation

Substitute 9.30 x 10¹³ Hz and c=10¹⁰ cm/sec in equation (III),

$$\begin{gathered} \overline{v}=\frac{9.30\times {10}^{13}\mathrm{Hz}}{3\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =\frac{9.30}{3}\times {10}^3{\mathrm{cm}}^{-1} \\ =3.10\times {10}^3{\mathrm{cm}}^{-1} \end{gathered}$$

Use this wave number and v_M = 1 cm/sec in the equation (IV),

$$\begin{gathered} f=\frac{2\left(1.00{\mathrm{cms}}^{-1}\right)\left(9.30\times {10}^{13}\mathrm{Hz}\right)}{3.00\times {10}^{10}{\mathrm{cms}}^{-1}} \\ =6.20\times {10}^3{\mathrm{s}}^{-1}\left(\frac{1\mathrm{kHz}}{{10}^3{\mathrm{s}}^{-1}}\right) \\ =6.20\mathrm{kHz} \end{gathered}$$