Question

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

Short answer

Frequency: $5.77\times {10}^{14}$ Hz, Energy: $3.82\times {10}^{-19}$ J or 2.39 eV, Momentum: $1.27\times {10}^{-27}$ kg⋅m/s.

Step-by-step solution

Identify Given Values

We are given that the wavelength of green light $\lambda =520\text{ nm}$. Remember to convert this to meters for calculations: $\lambda =520\times {10}^{-9}\text{ m}$.

Use the Speed of Light Formula to Find Frequency

The formula to relate speed of light $c$, wavelength $\lambda$, and frequency $f$ is $c=f\cdot \lambda$. Here, the speed of light $c=3.00\times {10}^8\text{ m/s}$. Solve for frequency $f$:$$f=\frac{c}{\lambda }=\frac{3.00\times {10}^8\text{ m/s}}{520\times {10}^{-9}\text{ m}}=5.77\times {10}^{14}\text{ Hz}$$

Calculate the Energy of the Photon in Joules

Use Planck's equation $E=h\cdot f$, where Planck's constant $h=6.63\times {10}^{-34}\text{ J}\cdot \text{s}$. Insert the frequency from Step 2:$$E=6.63\times {10}^{-34}\text{ J}\cdot \text{s}\times 5.77\times {10}^{14}\text{ Hz}=3.82\times {10}^{-19}\text{ J}$$

Convert the Energy from Joules to Electron Volts

1 electron volt $(\text{eV})$ is equivalent to $1.60\times {10}^{-19}\text{ J}$. Convert the energy calculated in Step 3:$$E=\frac{3.82\times {10}^{-19}\text{ J}}{1.60\times {10}^{-19}\text{ J/eV}}=2.39\text{ eV}$$

Calculate the Photon’s Momentum

Use the momentum formula $p=\frac{h}{\lambda }$, where $h=6.63\times {10}^{-34}\text{ J}\cdot \text{s}$. Insert the values:$$p=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{520\times {10}^{-9}\text{ m}}=1.27\times {10}^{-27}\text{ kg}\cdot \text{m/s}$$

Key concepts

Wavelength

In physics, wavelength is the distance between consecutive crests or troughs of a wave. For light or any electromagnetic wave moving through a vacuum, the speed is constant, and that's the speed of light, denoted as $c$. For visible light such as green light, which this exercise deals with, the wavelength is typically measured in nanometers (nm), which is a billionth of a meter.

• Green light has a wavelength of about 520 nm, which is 520 x 10^-9 meters when converted to standard scientific units.


• Wavelength is inversely proportional to frequency, meaning the longer the wavelength, the lower the frequency.

A key formula involving wavelength, frequency $f$, and the speed of light $c$ is $c=f\cdot \lambda$. By rearranging this formula, you can find the frequency if the wavelength is known, as was done in the original solution.

Frequency

Frequency $f$ is a measure of how often the waves of light (or any electromagnetic radiation) pass a point in space per second. It is measured in Hertz (Hz), with one Hertz equivalent to one cycle per second.

• For the photon of green light with a wavelength of 520 nm, the frequency was calculated using the formula $f=\frac{c}{\lambda }$.


• The speed of light $c$ is approximately $3.00\times {10}^8\text{ m/s}$, allowing us to calculate the frequency as $5.77\times {10}^{14}\text{ Hz}$.

It's important to understand that frequency is part of what defines the energy of a photon. According to Planck's equation $E=h\cdot f$, where $h$ is Planck's constant ($6.63\times {10}^{-34}\text{ J}\cdot \text{s}$), the energy is directly proportional to the frequency.

Momentum

Momentum is a concept typically used in contexts involving motion, mass, and speed. But for photons, which are massless particles of light, momentum is defined using a different approach. The momentum $p$ of a photon can be calculated using the formula $p=\frac{h}{\lambda }$, where $h$ is Planck's constant and $\lambda$ is the wavelength of the photon.

• This equation shows that even though photons have no mass, they possess momentum because they have energy.


• Thus, the greener light photons, with a wavelength of 520 nm ($520\times {10}^{-9}\text{ m}$), has a momentum of $1.27\times {10}^{-27}\text{ kg}\cdot \text{m/s}$.

Considering the small magnitude of a photon's momentum, it illustrates why light, despite being powerful, doesn't exert any perceivable push when it hits objects in our day-to-day experiences.