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
The Correct Answer and Explanation is:
To find the frequency, momentum, and energy of a green light photon with a wavelength of 520 nm, we’ll use fundamental equations from physics.
Given:
- Wavelength, λ=520\lambda = 520 nm = 520×10−9520 \times 10^{-9} m
- Speed of light, c=3.00×108c = 3.00 \times 10^8 m/s
- Planck’s constant, h=6.626×10−34h = 6.626 \times 10^{-34} J·s
- Electron charge, e=1.602×10−19e = 1.602 \times 10^{-19} C
1. Frequency ff:
f=cλ=3.00×108520×10−9=5.77×1014 Hzf = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{520 \times 10^{-9}} = 5.77 \times 10^{14} \text{ Hz}
2. Momentum pp:
p=hλ=6.626×10−34520×10−9=1.274×10−27 kg\cdotpm/sp = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{520 \times 10^{-9}} = 1.274 \times 10^{-27} \text{ kg·m/s}
3. Energy EE:
In joules:
E=hf=(6.626×10−34)(5.77×1014)=3.821×10−19 JE = hf = (6.626 \times 10^{-34})(5.77 \times 10^{14}) = 3.821 \times 10^{-19} \text{ J}
In electron volts (eV):
E=3.821×10−191.602×10−19=2.38 eVE = \frac{3.821 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.38 \text{ eV}
Explanation
Light exhibits both wave-like and particle-like properties. A photon, the quantum particle of light, carries energy and momentum despite having no mass. To understand its behavior, we use the wave-particle duality framework. In this case, we are given the wavelength of a green light photon as 520 nm (nanometers), which falls within the visible spectrum.
The frequency of the photon is calculated using the relation f=cλf = \frac{c}{\lambda}, where cc is the speed of light and λ\lambda is the wavelength. A shorter wavelength means a higher frequency, indicating more energetic photons. For 520 nm, the frequency turns out to be approximately 5.77×10145.77 \times 10^{14} Hz.
Next, using Planck’s relation E=hfE = hf, we calculate the energy carried by a single photon. Planck’s constant hh connects the wave (frequency) and particle (energy) nature of light. The resulting energy is 3.821×10−193.821 \times 10^{-19} joules, a very small value, as expected for individual photons. However, when dealing with large numbers of photons (like in a laser beam), the cumulative energy becomes significant. To express this energy in a more practical unit for atomic and quantum processes, we convert joules to electron volts, yielding 2.38 eV.
Finally, we compute the momentum using p=hλp = \frac{h}{\lambda}, which reflects that even massless particles like photons have momentum, critical in understanding phenomena like radiation pressure and the photoelectric effect.
These calculations show how fundamental constants link the electromagnetic nature of light with quantum behavior, reinforcing the photon model of light.
