Question

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in \(1859 .\) The name came from the Latin caesius ("sky blue") because of the prominent blue line observed for this element at \(455.5 \mathrm{~nm} .\) Calculate the frequency and energy of a photon of this light.

Short answer

The frequency of the photon of light with a wavelength of 455.5 nm is approximately \(6.58 \times 10^{14} \mathrm{Hz}\), and its energy is approximately \(4.36 \times 10^{-19} \mathrm{J}\).

Step-by-step solution

Convert the wavelength to meters

First, we need to convert the given wavelength from nanometers to meters, as this is the unit that we need for further calculations. To do this, we can use the conversion factor: \(1 \text{ nm} = 10^{-9} \text{ m}\) So, the wavelength in meters is: \(λ = 455.5 \times 10^{-9} \mathrm{m}\) Step 2: Calculate the frequency using the speed of light equation

Calculate the frequency using the speed of light equation

The speed of light equation is given as follows: \(c = λν\) where \(c\) is the speed of light (approximately \(3 \times 10^8 \mathrm{m/s}\)), λ is the wavelength (in meters), and \(ν\) is the frequency. Using the given wavelength, we can solve for the frequency: \(ν = \frac{c}{λ} = \frac{3 \times 10^8 \mathrm{m/s}}{455.5 \times 10^{-9} \mathrm{m}}\) Step 3: Calculate the value of the frequency

Calculate the value of the frequency

Using a calculator, we can find the value of the frequency: \(ν = 6.58 \times 10^{14} \mathrm{Hz}\) So, the frequency of the photon of light is \(6.58 \times 10^{14} \mathrm{Hz}\). Step 4: Calculate the energy using Planck's equation

Calculate the energy using Planck's equation

Planck's equation, which relates the energy of a photon to its frequency, is given as follows: \(E = hν\) where \(E\) is the energy of the photon, \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \mathrm{Js}\)), and \(ν\) is the frequency. Using the calculated frequency, we can find the energy of the photon: \(E = (6.63 \times 10^{-34} \mathrm{Js}) (6.58 \times 10^{14} \mathrm{Hz})\) Step 5: Calculate the value of the energy

Calculate the value of the energy

Using a calculator, we can find the value of the energy: \(E = 4.36 \times 10^{-19} \mathrm{J}\) So, the energy of the photon of light is \(4.36 \times 10^{-19} \mathrm{J}\). In conclusion, the frequency of the photon of light with a wavelength of 455.5 nm is approximately \(6.58 \times 10^{14} \mathrm{Hz}\), and its energy is approximately \(4.36 \times 10^{-19} \mathrm{J}\).

Key concepts

Wavelength Conversion

Let's start by understanding how to convert a wavelength from nanometers (nm) to meters (m). Wavelength is simply the distance between two peaks of a wave, like the peaks on water waves you might see at a beach. In scientific calculations, using the unit meter is preferred because it's the standard unit of length in the International System of Units (SI).

To convert nanometers to meters, we utilize the conversion factor:

• 1 nm = 10^-9 m

So, if you have a wavelength measured in nanometers, like 455.5 nm, you can convert it to meters by multiplying by 10^-9.

For instance: \[\lambda = 455.5 \times 10^{-9} \text{ m}\]
This conversion is essential as it allows us to use the wavelength in further calculations like frequency and energy, which are explained as follows.

Frequency Calculation

Understanding frequency is crucial in comprehending the nature of light. Frequency refers to how often the wave peaks pass a certain point over a period of time. In simpler terms, it's how "often" the waves appear. The frequency of light is directly related to its energy; a higher frequency means more energy.

To calculate the frequency, we use the formula:

• \( c = \lambda u \)

Here, \(c\) represents the speed of light (approximately \(3 \times 10^8\) m/s), \(\lambda\) is the wavelength in meters, and \(u\) is the frequency.

Rearranging the formula to solve for frequency gives us:

• \( u = \frac{c}{\lambda} \)

Applying the values, you substitute \(c\) and the converted \(\lambda\):\[u = \frac{3 \times 10^8\ \text{m/s}}{455.5 \times 10^{-9}\ \text{m}}\]
This yields a frequency of approximately \(6.58 \times 10^{14}\) Hz, indicating how frequently the light waves peak in a second.

Planck's Equation

Now, let's dive into how we calculate the energy of a photon using its frequency. Photons are tiny particles of light, and their energy can be calculated using **Planck's equation**:

• \( E = hu \)

In this equation, \(E\) is the energy of the photon, \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \) Js), and \(u\) is the frequency of the light.

By substituting the frequency you've calculated:\[E = (6.63 \times 10^{-34}\ \text{Js})(6.58 \times 10^{14}\ \text{Hz})\]
This calculation provides the energy of a single photon of light with the given frequency, leading to an energy of approximately \(4.36 \times 10^{-19}\) Joules.

Planck's equation helps bridge the wave properties of light with its particle properties, enabling a deeper understanding of how light functions both energetically and spatially.