Chapter 11, Problem 61A

Interpretation Introduction

Interpretation: The wave with higher wavelength, frequency, energy, and velocity needs to be identified. The values wavelength, frequency, and energy needs to be determined for each wave.

Concept introduction: Wavelength is defined as the gap between two successive crests or troughs of a wave. 

Relation between wavelength and frequency,

  $\begin{array}{l}\upsilon =\frac{c}{\lambda }\text{ where, }\upsilon =\text{ frequency}\\ c=\text{ speed of light}\\ \lambda =\text{ wavelength}\end{array}$

Relation between energy and frequency,

  $\begin{array}{l}E=h\upsilon \text{ where, }h=\text{ planck's constant}\\ E=\text{energy}\end{array}$

Velocity of the wave is $V=\upsilon \times \lambda$

Answer to Problem 61A

• Wave “a” is longer in wavelength and the value is $4\times {10}^{-4}m$ .
• Wave “b” has higher frequency and the value is $1.5\times {10}^{12}{s}^{-1}$ . Wave “b” has higher energy and the value is $9.9\times {10}^{-22}J$ .
• Both waves have same velocity.
• Both waves belong to infrared electromagnetic radiation.

Explanation of Solution

• Wave “a” is longer in wavelength as crests to crests distance is longer in it. Since there are four successive crests to crests gap wavelength is,

  $\begin{array}{l}\lambda =\frac{1.6\times {10}^{-3}m}{4}\\ \lambda =4\times {10}^{-4}m\end{array}$

• Wave “b” has lower wavelength. Since frequency is inversely proportional to wavelength Wave “b” has higher frequency.

Since energy is proportional to frequency energy of wave “b” has higher value.

In wave “b” there are 8 successive crests to crests gap and hence wavelength of this wave is,

  $\begin{array}{l}\lambda =\frac{1.6\times {10}^{-3}m}{8}\\ \lambda =2\times {10}^{-4}m\end{array}$

Frequency of this wave $\upsilon =\frac{3\times {10}^{8}m\cdot s^{-1}}{2\times {10}^{-4}m}=1.5\times {10}^{12}{s}^{-1}$

Energy of this wave $E=\left(6.626\times {10}^{-34}J\cdot s^{-1}\right)\times (1.5\times {10}^{12}{s}^{-1})=9.9\times {10}^{-22}J$

• Both waves have same velocity which is the speed of light $3\times {10}^{8}m\cdot s^{-1}$ .
• Range of infrared wavelength is ${10}^{-3}$ to ${10}^{-6}$ m. Since wavelength of both waves comes in this range they both belong to infrared electromagnetic radiation.

Conclusion

• Wave “a” is longer in wavelength and the value is $4\times {10}^{-4}m$ .
• Wave “b” has higher frequency and the value is $1.5\times {10}^{12}{s}^{-1}$ . Wave “b” has higher energy and the value is $9.9\times {10}^{-22}J$ .
• Both waves have same velocity.
• Both waves belong to infrared electromagnetic radiation.