Consider a single photon of red laser light, with wavelength of 633 nm (nano is 10-9).
What energy does this carry? What momentum? How does the energy compare to
typical atomic energies of a few times 10-19 J?
The energy is E = h*c/(wavelength)
The momentum is E/c = h/(wavelength)
h is Planck's constant, which you should know or look up. It has units of energy*time
The energy in a photon at that wavelength should be a few times 10^-19 J, or a few electron volts.
To calculate the energy of a photon, you can use the equation:
Energy = Planck's constant * Speed of light / Wavelength
Given:
Wavelength (λ) = 633 nm = 633 * 10^(-9) m
Planck's constant (h) = 6.626 * 10^(-34) Js
Speed of light (c) = 3 * 10^8 m/s
Plugging in the values:
Energy = (6.626 * 10^(-34) Js) * (3 * 10^8 m/s) / (633 * 10^(-9) m)
Calculating this:
Energy ≈ 9.87 * 10^(-19) J (joules)
So, the energy carried by a single photon of red laser light with a wavelength of 633 nm is approximately 9.87 * 10^(-19) J.
To calculate the momentum of a photon, you can use the equation:
Momentum = Planck's constant / Wavelength
Plugging in the values:
Momentum = (6.626 * 10^(-34) Js) / (633 * 10^(-9) m)
Calculating this:
Momentum ≈ 1.05 * 10^(-27) kg m/s (kilogram meters per second)
The energy of a single photon of red laser light, which is approximately 9.87 * 10^(-19) J, is much larger than typical atomic energies of a few times 10^(-19) J.
To find the energy carried by a single photon of red laser light, we can use the equation:
E = hc/λ
Where E is the energy of the photon, h is the Planck's constant (approximately 6.63 x 10^-34 Joule seconds), c is the speed of light (approximately 3 x 10^8 meters per second), and λ is the wavelength of the light.
Plugging in the values gives us:
E = (6.63 x 10^-34 J s) x (3 x 10^8 m/s) / (633 x 10^-9 m)
E ≈ 3.14 x 10^-19 Joules
So, the energy carried by a single photon of red laser light with a wavelength of 633 nm is approximately 3.14 x 10^-19 Joules.
To find the momentum of the photon, we can use the equation:
p = E/c
Where p is the momentum of the photon, E is the energy of the photon, and c is the speed of light.
Plugging in the value of E we found earlier and the value of c:
p = (3.14 x 10^-19 J) / (3 x 10^8 m/s)
p ≈ 1.05 x 10^-27 kg m/s
So, the momentum of the photon is approximately 1.05 x 10^-27 kilogram meters per second.
In terms of typical atomic energies of a few times 10^-19 J, the energy carried by the photon is of the same magnitude. Atomic energies usually range from a few electronvolts (eV) to a few electronvolts multiplied by Avogadro's number, which is approximately 6.02 x 10^23. Hence, the answer is that the energy of the photon is on par with typical atomic energies.