Orders of magnitude 2: What is the energy of a photon of wavelength λ=780 nm. Please give the energy in K⋅kB units this time (to within 1%). This is an equivalent temperature unit: (Tequiv=E/kB), with the temperature in Kelvin (K), and kB is the Boltzmann constant.
Answer for Question 2
To find the energy of a photon of a given wavelength, we can use the relationship:
E = hc / λ
where:
E is the energy of the photon
h is the Planck's constant (6.62607015 x 10^-34 J⋅s)
c is the speed of light in a vacuum (299,792,458 m/s)
λ is the wavelength of the photon
First, let's convert the wavelength from nanometers (nm) to meters (m):
λ = 780 nm = 780 x 10^-9 m
Now, we can calculate the energy of the photon using the formula:
E = (6.62607015 x 10^-34 J⋅s * 299,792,458 m/s) / (780 x 10^-9 m)
E = 2.68281285 x 10^-19 J
To convert this energy into K⋅kB units, we need to divide by the Boltzmann constant, kB:
Tequiv = E / kB
kB = 1.380649 x 10^-23 J/K (Boltzmann constant)
Tequiv = (2.68281285 x 10^-19 J) / (1.380649 x 10^-23 J/K)
Tequiv = 1945 K
Therefore, the energy of a photon with a wavelength of 780 nm is approximately equivalent to 1945 K⋅kB units.