Question

A sodium flame has a characteristic yellow color due to emission of light of wavelength \(589 \mathrm{nm}\). What is the mass equivalence of one photon with this wavelength \(\left(1 \mathrm{~J}=1 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\right) ?\)

Short answer

The mass equivalence of one photon with a wavelength of 589 nm is \( 3.751 \times 10^{-36} \text{kg} \).

Step-by-step solution

Find energy of the photon

Use the formula for the energy of a photon: \[ E = \frac{hc}{\lambda} \] where \[ h = 6.626 \times 10^{-34} \, \text{Js} \] is Planck's constant, \[ c = 3 \times 10^{8} \mathrm{~m/s} \] is the speed of light, and \[ \lambda = 589 \times 10^{-9} \mathrm{~m} \] is the wavelength. Substitute the values to get:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}} \]Calculate this to find the energy, \( E \).

Calculate energy

First, multiply the terms in the numerator:\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \]Next, divide by the wavelength:\[ \frac{1.9878 \times 10^{-25}}{589 \times 10^{-9}} = 3.376 \times 10^{-19} \text{J} \]Therefore, the energy of one photon is \( 3.376 \times 10^{-19} \text{J} \).

Use Einstein's mass-energy equivalence

Utilize Einstein's formula to find mass equivalence: \[ E = mc^2 \] Rearrange to solve for mass, \( m \): \[ m = \frac{E}{c^2} \]Substitute \( E = 3.376 \times 10^{-19} \text{J} \) and \( c = 3 \times 10^8 \mathrm{~m/s} \), giving:\[ m = \frac{3.376 \times 10^{-19}}{(3 \times 10^8)^2} = \frac{3.376 \times 10^{-19}}{9 \times 10^{16}} = 3.751 \times 10^{-36} \text{kg} \]

Key concepts

Wavelength and Frequency

The wavelength and frequency of light are fundamental concepts in physics, especially in quantum mechanics and electrodynamics.
Wavelength (\textbackslash( \textbackslash \textlambda \textbackslash)) is the distance between two successive peaks of a wave.
Frequency (\textbackslash( f \textbackslash)) is the number of waves that pass a point in one second.
These two properties are inversely related by the formula:
\[ c = \textbackslash \textlambda f \]
where \( c \) is the speed of light. When wavelength increases, frequency decreases and vice-versa.
In our problem, the wavelength of the emitted light from sodium flame is 589 nm (nanometers), which is visible as yellow light.
To find the frequency from the wavelength, use:
\[ f = \textbackslash \frac{c}{\textbackslash \textlambda} \]
This relationship highlights how energy can be distributed over different wavelengths and frequencies, impacting the color and type of light observed.

Planck's Constant

Planck's constant (\textbackslash( h \textbackslash)) is a fundamental constant in quantum mechanics.
It is defined as the ratio of photon energy to its frequency:
\( E = hf \)
Here, \( h = 6.626 \textbackslash times 10^{-34} \) Js.
Planck's constant allows us to link a photon's energy with its wavelength.
In the given problem, we used the formula:
\( E = \textbackslash \frac{hc}{\textbackslash \textlambda} \)
This helps find the energy of a photon based on its wavelength.
Understanding Planck’s constant offers insight into the nature of quantized energy levels in atoms and molecules.

Einstein's Mass-Energy Equivalence

Einstein's mass-energy equivalence principle is represented by the famous equation:
\( E = mc^2 \)
Here, \( E \) represents energy, \( m \) is mass, and \( c \) is the speed of light in a vacuum.
This principle states that energy and mass are interchangeable; they are different forms of the same thing.
In this exercise, we calculated the mass equivalence of a photon with a known energy:
First, we used the energy calculated from the wavelength, based on Planck's formula.
Then, we rearranged Einstein’s equation to solve for the mass:
\[ m = \textbackslash \frac{E}{c^2} \]
This shows that even a tiny amount of energy is related to a very small mass, highlighting the incredible power contained within particles of light, and the interplay between mass and energy at quantum scales.