Question

The frequency of light wave of wavelength \(5000 \mathrm{~A}\) is \(\mathrm{Hz}\) (A) \(6 \times 10^{14}\) (B) \(1.5 \times 10^{-2}\) (C) \(1.5\) (D) \(6 \times 10^{1}\)

Short answer

The frequency of the light wave with wavelength \(5000 \mathrm{~A}\) is \(6 \times 10^{14} \mathrm{Hz}\).

Step-by-step solution

Convert the wavelength to meters

The wavelength is given in angstroms (\(5000 \mathrm{~A}\)), but to use the speed of light in meters per second, we need to convert the wavelength to meters. The conversion factor is \(1 \mathrm{~A} = 10^{-10} \mathrm{m}\). \(λ = (5000 \mathrm{~A}) \times (10^{-10} \mathrm{m/A}) = 5 \times 10^{-7} \mathrm{m}\)

Calculate the frequency using the formula

Now that we have the wavelength (\(λ\)) in meters, we can use the formula \(c = λν\) to find the frequency (\(ν\)): \(ν = \frac{c}{λ}\) Since the value of \(c\) is \(3 \times 10^8 \mathrm{m/s}\) and the value of \(λ\) we calculated is \(5 \times 10^{-7} \mathrm{m}\), we can substitute these values and solve for \(ν\): \(ν = \frac{3 \times 10^8 \mathrm{m/s}}{5 \times 10^{-7} \mathrm{m}} = 6 \times 10^{14} \mathrm{Hz}\)

Choose the correct answer

The calculated frequency is \(6 \times 10^{14} \mathrm{Hz}\). Therefore, the correct answer is: (A) \(6 \times 10^{14}\)

Key concepts

Wavelength Conversion

When dealing with light waves, it's common to encounter wavelengths expressed in different units. In this problem, the wavelength is given in angstroms (\( \, \text{Å} \,\)). Angstroms are a convenient unit for atomic and molecular scales, but for calculations involving the speed of light, we must convert the wavelength to meters.
To convert angstroms to meters, remember that \(1 \, \text{Å} = 10^{-10} \, \text{m} \). So, for a wavelength of \(5000 \, \text{Å}\), the conversion to meters is:

\[ \lambda = 5000 \, \text{Å} \times 10^{-10} \, \text{m/Å} = 5 \times 10^{-7} \, \text{m} \]
This straightforward conversion is the foundation for further calculations with wavelengths in physics. Keeping track of units is crucial and using the correct conversion factor helps avoid errors.

Speed of Light

The speed of light (\( c \)) is a fundamental constant in physics, representing the speed at which light travels in a vacuum. It is approximately \(3 \times 10^8 \, \text{m/s} \). This constant plays a pivotal role in various equations, particularly those involving electromagnetic waves.

Understanding this concept is essential for problems involving the propagation of light and other electromagnetic phenomena. Because light travels incredibly fast, its speed allows us to calculate important properties of waves, such as frequency and energy. It's good to remember this constant as it frequently appears in physics calculations, ensuring quick and accurate problem-solving.

Frequency Formula

The relationship between the speed of light, wavelength, and frequency is given by the formula:

\[ c = \lambda u \]
where \(\lambda\) is the wavelength in meters and \(u\) is the frequency in hertz (\(\text{Hz}\)).
To find the frequency, rearrange the formula to:

\[ u = \frac{c}{\lambda} \]
This formula tells us that frequency is inversely proportional to wavelength, meaning that if the wavelength increases, the frequency decreases, and vice versa.

This inverse relationship is a key idea in wave physics, showing how energy spreads over a wave. Using known values of \(c\) and \(\lambda\), we can easily calculate \(u\), applying it to various scenarios involving light and electromagnetic waves.

Physics Problem Solving

Solving physics problems systematically ensures accuracy and understanding. Begin by identifying what you know and what you need to find. In this problem, given the wavelength in angstroms, the goal was to find the frequency of the light wave.
Here's a step-by-step approach:

• Convert units as needed, ensuring all measurements align with known constants.
• Apply the correct formula, here it was \( u = \frac{c}{\lambda} \).
• Substitute the values you know into the formula.
• Calculate the result and double-check with any provided options, if applicable.

This methodical process helps not only in reaching the correct solution but also in building a deeper understanding of the underlying physical principles.