Chapter 5, Problem 50A

Interpretation Introduction

Interpretation:

The wavelength of the orange line from the figure needs to be determined and also determine the frequency and energy of the photon.

Concept introduction:

A photon is a massless particle conveying a quantum of energy. Energy of a photon is calculated by formula $E=\text{hν}$. Here, E represents energy, $\text{h}$ and $\text{ν}$ represents Planck’s constant and frequency respectively.

Answer to Problem 50A

Frequency is $4.88\times {10}^{14}{\text{s}}^{-1}$ and energy of the photon is $\text{30}\text{.33}\times {10}^{-20}\text{J}$.

Explanation of Solution

From the given figure, the wavelength is approximately $615\text{nm}$.

The speed of an electromagnetic wave is expressed by below formula:

$\begin{array}{l}\text{Speed}=\text{wavelength}\times \text{frequency}\\ \text{frequency}=\frac{\text{Speed}}{\text{wavelength}}\end{array}$ …… (1)

Convert nanometer into meter as follows:

$\begin{array}{c}1\text{nm}={10}^{-9}\text{m}\\ \text{615}\text{nm}=\frac{\text{615}\text{nm}\times {10}^{-9}\text{m}}{1\text{nm}}\\ =6.15\times {10}^{-7}\text{m}\end{array}$

The value of speed is $3\times {10}^8\text{m}/\text{s}$.

Substitute above values in equation (1).

$\begin{array}{c}\text{frequency}=\frac{\text{Speed}}{\text{wavelength}}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{6.15\times {10}^{-7}\text{m}}\\ =0.488\times {10}^{8+7}{\text{s}}^{-1}\\ =4.88\times {10}^{14}{\text{s}}^{-1}\end{array}$

Hence, frequency is $4.88\times {10}^{14}{\text{s}}^{-1}$.

Now, calculate energy of the photon by the expression given below:

$E=\text{hν}$ …… (2)

The value of h is $6.626\times {10}^{-34}\text{J}.\text{s}$.

Substitute the value of h and frequency in equation (2) as follows:

$\begin{array}{c}E=\text{hν}\\ \text{=4}\text{.88}\times {\text{10}}^{14}{\text{s}}^{-1}\times 6.626\times {10}^{-34}\text{J}.\text{s}\\ \text{=30}\text{.33}\times {10}^{-20}\text{J}\end{array}$

Hence, energy is $\text{30}\text{.33}\times {10}^{-20}\text{J}$.

Conclusion

Frequency is $4.88\times {10}^{14}{\text{s}}^{-1}$ and energy of the photon is $\text{30}\text{.33}\times {10}^{-20}\text{J}$.