Chapter 7, Problem 1PE

Helium derives its name from the Latin name for the sun. Helium was discovered in 1868 when spectroscopists found that the 588 nm wavelength (among others) was missing from the sun's spectrum. What is the frequency of this radiation? (Hint: Recall the metric prefixes so units will cancel correctly.)

Interpretation Introduction

Interpretation:

The frequency of wave of $588\times {10}^{-9}\text{ m}$ wavelength that is missing from the Sun’s spectrum is to be determined.

Concept Introduction:

Frequency of a wave is defined as the number of waves that pass through a fixed point in a given time period. It is denoted by $\text{ν}$ and is measured in hertz $\left(\text{Hz}\right)$.

The wavelength and frequency of a wave are correlated by the general formula which is written as:

$\nu =\frac{\text{c}}{\lambda }$,

Here, $\text{c}$ is the speed of light and $\lambda$ is the wavelength of the wave.

Answer to Problem 1PE

Solution: $5.10\times {10}^{14}\text{ Hz}$

Explanation of Solution

The frequency and wavelength of a wave are expressed by the following equation:

$\nu =\frac{\text{c}}{\lambda }$

Here, $\text{c}$ is the speed of light that is, $3\times {10}^8\text{ m/s}$ and $\lambda$ is the wavelength of the wave.

Substitute, $3\times {10}^8\text{ m/s}$ for speed of light and $588\times {10}^{-9}\text{ m}$ for wavelength in the above equation as follows:

$\begin{array}{c}\nu =\left(\frac{3\times {10}^8\text{ m/s}}{588\times {10}^{-9}\text{ m}}\right)\\ =5.10\times {10}^{14}\text{ Hz}\end{array}$

Conclusion

Frequency of the wave of wavelength $588\text{ nm}$ that is missing from the Sun’s spectrum is $5.10\times {10}^{14}\text{ Hz}$.