Chapter 6, Problem 20QP

A particular form of electromagnetic radiation has a frequency of $\text{9}\text{.87 }\times {\text{ 10}}^{\text{15}}$ Hz. (a) What is its wavelength in nanometers? In meters? (b) To what region of the electromagnetic spectrum would you assign it? (c) What is the energy (in joules) of one quantum of this radiation?

Interpretation Introduction

Interpretation:

The wavelength of the photon, the region of the electromagnetic spectrum, and the energy (in joules) for one quantum of this radiation is to be determined.

Concept introduction:

The range of wavelengths of electromagnetic radiations and their respective frequencies all fall under electromagnetic spectrum. The range of wavelengths (in nm) and the types of radiation are as follows: ${\text{10}}^{-3}\text{ to }{10}^{-1}$—gamma, ${10}^{-1}\text{ to 10}$—X-ray, $10{\text{ to 10}}^3$—ultraviolet, 400 to 700—visible, ${\text{10}}^3{\text{ to 10}}^5$—infrared, ${\text{10}}^5{\text{ to 10}}^7$—microwave, and ${\text{10}}^7{\text{ to 10}}^{13}$—radio wave.

The relationship between wavelength and frequency of an electromagnetic radiation is as follows:

$\text{c}=\lambda \times \nu$

Here, $\text{c}$ is the speed of light $\left(3.00\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is frequency.

The energy of a photon can be expressed as follows:

$\begin{array}{c}E=\text{h}\nu \\ \text{or, }E=\frac{\text{hc}}{\lambda }\end{array}$

Here, E is the energy of photon, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is frequency.

Conversion of meter ($\text{m}$) to nanometer $\text{nm}$ is $\frac{1.0\text{ nm}}{1.0\times {10}^{-9}\text{ m}}$.

Conversion of $\text{Hz}$ to per second is $\frac{1.0\text{ Hz}}{1.0{\text{ s}}^{-1}}$.

Answer to Problem 20QP

Solution:

(a)

$30.4\text{ nm}$ and $3.04\times {10}^{-8}\text{ m}$

(b)

Ultraviolet region

(c)

$6.54\times {10}^{-18}\text{ J}$

Explanation of Solution

Given information: $\nu =9.87\times {10}^{15}\text{ Hz}$

a) Wavelength of electromagnetic radiation in nanometer

The wavelength (in nm) of radiation can be evaluated as follows:

$\lambda =\frac{\text{c}}{\nu }$

Substitute $3.00\times {10}^8\text{ m/s}$ for $\text{c}$ and $9.87\times {10}^{15}\text{ Hz}$ for $\nu$

$\begin{array}{c}\lambda =\frac{3.0\times {10}^8\cancel{\text{m}}\text{.}\cancel{{\text{s}}^{-1}}}{9.87\times {10}^{15}\cancel{\text{Hz}}}\left(\frac{1.0\cancel{\text{Hz}}}{1.0\cancel{{\text{s}}^{-1}}}\right)\left(\frac{1.0\text{ nm}}{1.0\times {10}^{-9}\cancel{\text{m}}}\right)\\ =30.4\text{ nm}\end{array}$

Therefore, the wavelength of the radiation is $30.4\text{ nm}$.

The wavelength (in nm) of the radiation can be converted into meters as follows:

$\begin{array}{c}\lambda \left(\text{in m}\right)=\lambda \left(\text{in nm}\right)\left(\frac{1.0\times {10}^{-9}\text{ m}}{\text{1}\text{.0 nm}}\right)\\ =\left(30\cancel{\text{nm}}\right)\left(\frac{1.0\times {10}^{-9}\text{ m}}{\text{1}\text{.0 }\cancel{\text{nm}}}\right)\\ =3.04\times {10}^{-8}\text{ m}\end{array}$

Therefore, the wavelength (in m) of the radiation is $3.04\times {10}^{-8}\text{ m}$.

b) The region of the electromagnetic spectrum

The wavelength range of ultraviolet region is from $10\text{ nm}$ to $400\text{ nm}$. The wavelength of radiation is $30.4\text{ nm}$. Therefore, this radiation falls in the ultraviolet region.

c) The energy (in joules) of one quantum of electromagnetic radiation

The energy of a quantum of radiation is evaluated as shown below:

$E=\text{h}\nu$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\text{h}$ and $9.87\times {10}^{15}\text{ Hz}$ for $\nu$

$\begin{array}{c}E=\left(6.63\times {10}^{-34}\text{ J}\cdot \cancel{\text{s}}\right)\left(9.87\times {10}^{15}\cancel{\text{Hz}}\right)\left(\frac{1.0\cancel{{\text{s}}^{-1}}}{1.0\cancel{\text{Hz}}}\right)\\ =6.54\times {10}^{-18}\text{ J}\end{array}$

Hence, the energy of a quantum of radiation is $6.54\times {10}^{-18}\text{ J}$.