Chapter 8, Problem 13STP

Interpretation Introduction

Interpretation:

The frequency of the photon emitted by the given element is to be calculated.

Concept introduction:

When light passes through prism, it separates into different colors of light with different frequencies. The pattern of these different lines is termed as atomic emission spectrum.

Answer to Problem 13STP

The frequency of the photon emitted by the given element is $5.2\times {10}^{14}\text{Hz}$.

Explanation of Solution

The atomic emission spectrum of given element is shown below.

The above atomic emission spectrum of the given element shows that the first emission band appears at 580 nm. Therefore, the wavelength $\left(\lambda \right)$ of photon emitted by the given element via using atomic emission spectrum is 580 nm.

The frequency $\left(\upsilon \right)$ of photon is calculated by the formula shown below.

$\begin{array}{l}c=\lambda \upsilon \\ \upsilon =\frac{c}{\lambda }\end{array}$

Where,

• $\lambda$ is the wavelength.
• $c$ is the speed of light $\left(3.0\times {10}^8\text{m/s}\right)$ .

Substitute the given data in above equation.

$\begin{array}{c}\upsilon =\frac{c}{\lambda }\\ =\frac{3.0\times {10}^8\text{m/s}}{580\text{nm}}\\ =\frac{3.0\times {10}^8\text{m/s}}{580\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =\frac{3.0\times {10}^8\text{m/s}}{5.80\times {10}^{-7}\text{m}}\\ =0.52\times {10}^{15}{\text{s}}^{-1}\\ \upsilon =5.2\times {10}^{14}\text{Hz}\end{array}$

The frequency $\left(\upsilon \right)$ of photon is $5.2\times {10}^{14}\text{Hz}$.

Conclusion

The frequency of the photon emitted by the given element is $5.2\times {10}^{14}\text{Hz}$.