Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Answer to Problem 38Q

The energy of the waves will be arranged as:

$\begin{array}{l}10\times {10}^{-26}>6.63\times {10}^{-26}>3.3\times {10}^{-26}\\ II=VI>I=V>III=IV\end{array}$.

Explanation of Solution

Given data:

Wave I: $\lambda$ = 3m.

Wave II: $\lambda$ = 2m.

Wave III: $\lambda$ = 6m.

Wave IV: $\lambda$ = 6m.

Wave V: $\lambda$ = 3m.

Wave VI: $\lambda$ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

$\begin{array}{l}E=h\nu \\ c=\lambda \nu \end{array}$.

Calculation:

Having the known values, the energy of the wave is calculated as:

$\begin{array}{l}E=h\nu \\ c=\lambda \nu \end{array}$

Where, E is the energy of the wave.

h is the Planck’s constant; $h=6.63\times {10}^{-34}Js$

c is the velocity of light; $c=3\times {10}^{8}m{s}^{-1}$

$\begin{array}{l}\text{Wave I}:\lambda =3m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{3}={10}^8{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left({10}^8\right)=6.63\times {10}^{-26}J{s}^{-1}\\ \text{Wave II}:\lambda =2m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{2}=15\times {10}^7{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left(15\times {10}^7\right)=10\times {10}^{-26}J{s}^{-1}\\ \text{Wave III}:\lambda =6m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{6}=0.5\times {10}^8{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left(0.5\times {10}^8\right)=3.3\times {10}^{-26}J{s}^{-1}\\ \text{Wave IV}:\lambda =6m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{6}=0.5\times {10}^8{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left(0.5\times {10}^8\right)=3.3\times {10}^{-26}J{s}^{-1}\end{array}$

$\begin{array}{l}\text{Wave V}:\lambda =3m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{3}={10}^8{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left({10}^8\right)=6.63\times {10}^{-26}J{s}^{-1}\end{array}$

$\begin{array}{l}\text{Wave VI}:\lambda =2m\\ \nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{2}=15\times {10}^7{s}^{-1}\\ E=h\nu \\ E=(6.63\times {10}^{-34})\left(15\times {10}^7\right)=10\times {10}^{-26}J{s}^{-1}\end{array}$.

Conclusion:

Thus, the energy of the waves are arranged as:

$\begin{array}{l}10\times {10}^{-26}>6.63\times {10}^{-26}>3.3\times {10}^{-26}\\ II=VI>I=V>III=IV\end{array}$.