Chapter 18, Problem 18.1P

(a)

Interpretation Introduction

Interpretation:

Blank in the description of frequency of electromagnetic radiation has to be filled.

Concept Introduction:

Formula to calculate frequency is as follows:

  $v=\frac{c}{\lambda }$

Planck’s hypothesis states energy associated with photon is as follows:

  $\begin{array}{c}E=hv\\ =\frac{hc}{\lambda }\end{array}$

Formula to calculate wavenumber is as follows:

  $\overline{v}=\frac{v}{c}$

Here,

$\lambda$ denotes wavelength.

$h$ denotes Planck’s constant

$c$ denotes speed of light.

$\overline{v}$ denotes wavenumber.

Explanation of Solution

Planck’s hypothesis states energy associated with photon in terms of frequency is as follows:

  $E=hv$

Since energy is directly related to frequency in accordance with Planck’s hypothesis thus when frequency is doubled, energy is also doubled. So energy will also double.

(b)

Interpretation Introduction

Interpretation:

Blank in the description of wavelength and energy of electromagnetic radiation has to be filled.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Planck’s hypothesis states energy associated with photon in terms of wavelength is as follows:

  $E=\frac{hc}{\lambda }$

Since energy is inversely directly related to energy in accordance with Planck’s hypothesis thus when wavelength is doubled, energy is reduced by same amount doubled. So, energy will become halve.

(c)

Interpretation Introduction

Interpretation:

Blank in the description of wavenumber and energy of electromagnetic radiation has to be filled.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Planck’s hypothesis states energy associated with photon in terms of wavenumber is as follows:

  $E=hc\overline{v}$

Since energy is inversely directly related to $\overline{v}$ in accordance with Planck’s hypothesis thus when $\overline{v}$ is doubled, energy is increased by same amount. So, energy will also become double.