What is the wavelength of a yellow light with a frequency of [tex]$5.2 \times 10^{14} Hz$[/tex]? Give your answer in proper scientific notation.
[tex]
\begin{array}{l}
{[?] \times 10^{[?]} m} \\
c=3.0 \times 10^8 m / s
\end{array}
[/tex]
Answer (1)
Use the formula c = λ f to relate the speed of light, wavelength, and frequency.
Rearrange the formula to solve for the wavelength: λ = f c .
Substitute the given values: λ = 5.2 × 1 0 14 Hz 3.0 × 1 0 8 m / s .
Calculate the wavelength and express the result in scientific notation: λ = 5.769 × 1 0 − 7 m .
Explanation
Understanding the Problem We are asked to find the wavelength of yellow light given its frequency. We know that the speed of light, frequency, and wavelength are related by the formula: c = λ f where c is the speed of light, λ is the wavelength, and f is the frequency.
Identify Given Information We are given:
Frequency f = 5.2 × 1 0 14 Hz
Speed of light c = 3.0 × 1 0 8 m / s We want to find the wavelength λ .
Rearrange the Formula We can rearrange the formula to solve for the wavelength: λ = f c
Substitute the Values Now, we substitute the given values into the formula: λ = 5.2 × 1 0 14 Hz 3.0 × 1 0 8 m / s
Simplify the Expression We can simplify this expression: λ = 5.2 3.0 × 1 0 8 − 14 m λ = 5.2 3.0 × 1 0 − 6 m
Calculate the Division Now we divide 3.0 by 5.2: 5.2 3.0 ≈ 0.5769
Express in Scientific Notation So, the wavelength is approximately: λ = 0.5769 × 1 0 − 6 m To express this in proper scientific notation, we need to have the coefficient between 1 and 10. So we rewrite this as: λ = 5.769 × 1 0 − 7 m
Examples
Understanding the wavelength of light is crucial in various applications, such as designing optical instruments like microscopes and telescopes. For example, the resolution of a microscope is limited by the wavelength of light used; shorter wavelengths allow for finer details to be observed. In telecommunications, different wavelengths of light are used to transmit data through fiber optic cables. Knowing the wavelength helps engineers select the appropriate materials and components for efficient data transmission.
