Chapter 5, Problem 81A

a)

Interpretation Introduction

Interpretation : The frequencies of the given wavelengths 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 photon.

Answer to Problem 81A

The frequency ${\nu }_1$ is $\text{5}{\text{.20×10}}^{12}{\text{s}}^{-1}$ .

The frequency ${\nu }_2$ is $\text{4}{\text{.40×10}}^{13}{\text{s}}^{-1}$ .

The frequency ${\nu }_3$ is $\text{9}{\text{.49×10}}^{13}{\text{s}}^{-1}$ .

The frequency ${\nu }_4$ is $\text{1}{\text{.70×10}}^{14}{\text{s}}^{-1}$ .

The frequency ${\nu }_5$ is $\text{2}{\text{.20×10}}^{14}{\text{s}}^{-1}$ .

The frequency ${\nu }_6$ is $\text{4}{\text{.70×10}}^{14}{\text{s}}^{-1}$ .

Explanation of Solution

Given information:

The wavelength ${\lambda }_1$ is $\text{5}{\text{.77×10}}^{-\text{3}}\text{cm}$ .

The wavelength ${\lambda }_2$ is $\text{6}{\text{.82×10}}^{-4}\text{cm}$ .

The wavelength ${\lambda }_3$ is $\text{3}{\text{.16×10}}^{-4}\text{cm}$ .

The wavelength ${\lambda }_4$ is $\text{1}{\text{.76×10}}^{-4}\text{cm}$ .

The wavelength ${\lambda }_5$ is $\text{1}{\text{.36×10}}^{-4}\text{cm}$ .

The wavelength ${\lambda }_6$ is $\text{6}{\text{.38×10}}^{-5}\text{cm}$ .

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 ${\nu }_1$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_1=\frac{c}{{\lambda }_1}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{5.77\times {10}^{-3}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =5.20\times {10}^{12}{\text{s}}^{-1}\end{array}$

To calculate the frequency ${\nu }_2$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_2=\frac{c}{{\lambda }_2}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{6.82\times {10}^{-4}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =4.40\times {10}^{13}{\text{s}}^{-1}\end{array}$

To calculate the frequency ${\nu }_3$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_3=\frac{c}{{\lambda }_3}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{3.16\times {10}^{-4}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =9.49\times {10}^{13}{\text{s}}^{-1}\end{array}$

To calculate the frequency ${\nu }_4$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_4=\frac{c}{{\lambda }_4}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{1.76\times {10}^{-4}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =1.70\times {10}^{14}{\text{s}}^{-1}\end{array}$

To calculate the frequency ${\nu }_5$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_5=\frac{c}{{\lambda }_5}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{1.36\times {10}^{-4}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =2.20\times {10}^{14}{\text{s}}^{-1}\end{array}$

To calculate the frequency ${\nu }_6$ , substitute the values in the above formula.

  $\begin{array}{c}{\nu }_6=\frac{c}{{\lambda }_6}\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{6.38\times {10}^{-5}\cancel{\text{cm}}}\times \frac{{10}^2\cancel{\text{cm}}}{1\cancel{\text{m}}}\\ =4.70\times {10}^{14}{\text{s}}^{-1}\end{array}$

b)

Interpretation Introduction

Interpretation : The energy of the photon and frequency are to be plotted.

Concept Introduction : Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v). The formula of energy is given as

  $E=h\nu$

Where $\text{h}$ is Planck’s constant and $\nu$ is the frequency of the photon.

Answer to Problem 81A

The energy of the photon is plotted on the y-axis while the frequency in the x-axis.

Explanation of Solution

The calculated frequencies and energy of the photon are tabulated.

Frequency $\left({10}^{12}{\text{s}}^{-1}\right)$	Energy of photon $\left({10}^{-19}\text{J}\right)$
5.20	345
44.0	29.2
94.90	62.9
170	1.13
220	1.46
470	3.11

The graph is plotted by frequency on the x-axis and energy of the photon on the y-axis. The plotting is done according to Planck’s energy formula.

  

c)

Interpretation Introduction

Interpretation : The energy of the photon and frequency is to be plotted.

Concept Introduction : Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v).

Answer to Problem 81A

The slope of the line obtained is $6.6\times {10}^{-34}\text{Js}$ .

Explanation of Solution

From the graph, the obtained equation is $y=6.6\times {10}^{-34}x$ .

According to the equation $y=mx$ , the value of the slope is $6.6\times {10}^{-34}$ . The value of the slope with the unit is $6.6\times {10}^{-34}\text{Js}$ .

d)

Interpretation Introduction

Interpretation : The energy of the photon and frequency is to be plotted.

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 81A

The slope of the line obtained is Planck’s constant.

Explanation of Solution

The significance of the slope is Planck’s constant.

Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v). The formula of energy is given as

  $E=h\nu$

Where $\text{h}$ is Planck’s constant and $\nu$ is the frequency of the photon.