Chapter 5, Problem 64A

Interpretation Introduction

Interpretation : The wavelengths of the radiation in centimeters are to be calculated.

Concept Introduction : The separation between the crests is known as the wavelength.

The frequency is how often waves cycle through a specific point in a unit of time.

Answer to Problem 64A

The wavelengths of the radiation in centimeters are $5.890\times {10}^{-5}\text{cm}$ and $5.896\times {10}^{-5}\text{cm}$ .

Explanation of Solution

Given information:

The wavelengths are $5.890\times {10}^{-7}\text{m}$ and $5.896\times {10}^{-7}\text{m}$ .

It is known $1\text{m}\text{=}{\text{10}}^2\text{cm}$ .

The separation between the crests is known as the wavelength.

The frequency is how often waves cycle through a specific point in a unit of time.

With the help of unit conversion from meter to centimeter, the wavelengths in centimeters are calculated.

The wavelength $5.890\times {10}^{-7}\text{m}$ is given as:

  $\begin{array}{c}\lambda =5.890\times {10}^{-7}\text{m}\\ =5.890\times {10}^{-7}\cancel{\text{m}}\times \frac{{10}^2\text{cm}}{1\cancel{\text{m}}}\\ \lambda =5.890\times {10}^{-5}\text{cm}\end{array}$

The wavelength $5.896\times {10}^{-7}\text{m}$ is given as:

  $\begin{array}{c}\lambda =5.896\times {10}^{-7}\text{m}\\ =5.896\times {10}^{-7}\cancel{\text{m}}\times \frac{{10}^2\text{cm}}{1\cancel{\text{m}}}\\ \lambda =5.896\times {10}^{-5}\text{cm}\end{array}$

The wavelengths of the radiation are $5.890\times {10}^{-5}\text{cm}$ and $5.896\times {10}^{-5}\text{cm}$ .

Interpretation Introduction

Interpretation : The frequencies of the radiations are to be calculated.

Concept Introduction : The separation between the crests is known as the wavelength.

The frequency is how often waves cycle through a specific point in a unit of time.

The formula of frequency $\left(\nu \right)$ is given as:

  $\nu =\frac{c}{\lambda }$

Where $c$ is the velocity of light and $\lambda$ is the wavelength of the photon.

Answer to Problem 64A

The frequencies of the radiations are $5.090\times {10}^{14}\text{Hz}$ and $5.085\times {10}^{14}\text{Hz}$ .

Explanation of Solution

Given information:

The wavelengths are $5.890\times {10}^{-7}\text{m}$ and $5.896\times {10}^{-7}\text{m}$ .

It is known that velocity of light is $2.998\times {10}^8\text{m/s}$ .

The separation between the crests is known as the wavelength.

The frequency is how often waves cycle through a specific point in a unit of time.

The formula of frequency $\left(\nu \right)$ is given as:

  $\nu =\frac{c}{\lambda }$

Where $c$ is the velocity of light and $\lambda$ is the wavelength of the photon.

To calculate the frequency of $5.890\times {10}^{-7}\text{m}$ radiation, substitute the values in the above formula.

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{5.890\times {10}^{-7}\cancel{\text{m}}}\\ =0.5090\times {10}^{15}{\text{s}}^{-1}\\ \nu =5.090\times {10}^{14}\text{Hz}\end{array}$

To calculate the frequency of $5.896\times {10}^{-7}\text{m}$ radiation, substitute the values in the above formula.

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{5.896\times {10}^{-7}\cancel{\text{m}}}\\ =0.5085\times {10}^{15}{\text{s}}^{-1}\\ \nu =5.085\times {10}^{14}\text{Hz}\end{array}$

The frequencies of the radiations are $5.090\times {10}^{14}\text{Hz}$ and $5.085\times {10}^{14}\text{Hz}$ .

Interpretation Introduction

Interpretation : The region of the electromagnetic spectrum is to be identified.

Concept Introduction : A wide range of wavelengths of radiation make up the electromagnetic spectrum. It has a wavelength in the ${10}^{-7}\text{m}$ range and a frequency in the $\text{1015 Hz}$ range.

Answer to Problem 64A

The region of the electromagnetic spectrum lies in the visible region.

Explanation of Solution

A wide range of wavelengths of radiation makes up the electromagnetic spectrum.

It has a wavelength in the ${10}^{-7}\text{m}$ range and a frequency in the $\text{1015 Hz}$ range.

To find the region of the electromagnetic spectrum, the wavelength should be expressed in nanometers.

The wavelength $5.890\times {10}^{-7}\text{m}$

(in nanometer) is given as:

  $\begin{array}{c}\lambda =5.890\times {10}^{-7}\text{m}\\ =5.890\times {10}^{-7}\cancel{\text{m}}\times \frac{{10}^9\text{nm}}{1\cancel{\text{m}}}\\ \lambda =5.89\times {10}^2\text{nm}\\ \text{= 589}\text{nm}\end{array}$

The wavelength $5.896\times {10}^{-7}\text{m}$

(in nanometer) is given as:

  $\begin{array}{c}\lambda =5.896\times {10}^{-7}\text{m}\\ =5.896\times {10}^{-7}\cancel{\text{m}}\times \frac{{10}^9\text{nm}}{1\cancel{\text{m}}}\\ =5.896\times {10}^2\text{nm}\\ \lambda \text{= 589}\text{.6}\text{nm}\end{array}$

Since the value of wavelength is low, it belongs to the visible region whose range is in between $380\text{nm}$ and $\text{700}\text{nm}$ .