Chapter 1, Problem 1A.10E

Interpretation Introduction

Interpretation:

The given table has to be completed.

Frequency	Wavelength	Energy of photon	Event
		$2.7\times {10}^{-19}J$
	$999nm$
$8.7\times {10}^{14}Hz$
	$155cm$

Concept Introduction:

The relationship between wavelength, frequency and speed of light is given below.

  $\lambda \times \nu =c$

Where, $\lambda$ is the wavelength, $\nu$ is the frequency of radiation and $c$ is the speed of light.

Energy of a photon can be expressed mathematically as given below.

  $E=h\nu =\frac{hc}{\lambda }$

The wavelengths of electromagnetic radiation and their corresponding frequencies are given below in the table.

Radiation type	Frequency / $\left({10}^{14}Hz\right)$	Wavelength / $\left(nm\right)$	Energy of photon / $\left({10}^{-19}J\right)$
x-rays and $\gamma$- rays	$\ge {10}^3$	$\le 3$	$\ge {10}^3$
Ultraviolet	$8.6$	$350$	$5.7$
Visible light
Violet	$7.1$	$420$	$4.7$
Blue	$6.4$	$470$	$4.2$
Green	$5.7$	$530$	$3.8$
Yellow	$5.2$	$580$	$3.4$
Orange	$4.8$	$620$	$3.2$
Red	$4.3$	$700$	$2.8$
Infrared	$3.0$	$1000$	$2.0$
Microwaves and radio waves	$\le {10}^{-3}$	$\ge 3\times {10}^6$	$\le {10}^3$

Explanation of Solution

First row:

The energy of the radiation is $2.7\times {10}^{-19}J.$

The calculation of wavelength and frequency of the radiation can be done as shown below.

  $\begin{array}{l}E=\frac{hc}{\lambda }\\ \\ \lambda =\frac{hc}{E}=\frac{\left(6.6\times {10}^{-34}J.s\right)\times \left(3\times {10}^{8}m/s\right)}{2.7\times {10}^{-19}J}=7.33\times {\text{10}}^{\text{-7}}m=733nm\\ \\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^{8}m/s}{733\times {10}^{-9}m}=4.09\times 10^{14}Hz\end{array}$

Second row:

The wavelength of the radiation is $999nm.$

The calculation of frequency and energy of the radiation can be done as shown below.

  $\begin{array}{l}\nu =\frac{c}{\lambda }=\frac{3\times {10}^{8}m/s}{999\times {10}^{-9}m}=3.00\times 10^{14}Hz\\ \\ E=h\nu =\left(6.6\times {10}^{-34}J.s\right)\times \left(\text{3}{\text{.00×10}}^{\text{14}}{s}^{-1}\right)=1.98\times 10^{-19}J\end{array}$

Third row:

The frequency of the radiation is $5\times {10}^{19}Hz.$

The calculation of wavelength and energy of the radiation can be done as shown below.

  $\begin{array}{l}\lambda =\frac{c}{\nu }=\frac{3\times {10}^{8}m/s}{5\times {10}^{19}{s}^{-1}}=6.0\times {10}^{-12}m=0.006nm\\ \\ E=\frac{hc}{\lambda }=\frac{\left(6.6\times {10}^{-34}J.s\right)\times \left(3\times {10}^{8}m/s\right)}{6.0\times {10}^{-12}m}=3.3\times 10^{-14}J\end{array}$

Fourth row:

The wavelength of the radiation is $155cm.$

The calculation of frequency and energy of the radiation can be done as shown below.

  $\begin{array}{l}\nu =\frac{c}{\lambda }=\frac{3\times {10}^{8}m/s}{155\times {10}^{-2}m}=0.02\times 10^{14}Hz\\ \\ E=h\nu =\left(6.6\times {10}^{-34}J.s\right)\times \left(\text{0}{\text{.02×10}}^{\text{14}}{s}^{-1}\right)=1.32\times 10^{-21}J\end{array}$

Now, the given table can be filled up as given below.

Frequency	Wavelength	Energy of photon	Event
$4.09\times 10^{14}Hz$	$733nm$	$2.7\times {10}^{-19}J$	Red traffic light
$3.00\times 10^{14}Hz$	$999nm$	$1.98\times 10^{-19}J$	Infrared lamp
$5\times {10}^{19}Hz$	$0.006nm$	$3.3\times 10^{-14}J$	$\gamma$- ray from outer space
$0.02\times 10^{14}Hz$	$155cm$	$1.32\times 10^{-21}J$	Listening to the car radio