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The photon flux per unit area is the photon irradiance if the photons are incident on a surface, or photon exitance if the emission of photons from a broad-area source is being considered. The flux per unit solid angle is the photon intensity. The flux per unit source area per unit solid angle is photon radiance. SI units for these quantities ...
The daily light integral (DLI) is the number of photosynthetically active photons (photons in the PAR range) accumulated in a square meter over the course of a day. It is a function of photosynthetic light intensity and duration (day length) and is usually expressed as moles of light (mol photons) per square meter (m −2) per day (d −1), or: mol·m −2 ·d −1.
count of photons n with energy Q p = h c/λ. [nb 2] photon flux: Φ q: count per second: s −1: T −1: photons per unit time, dn/dt with n = photon number. also called photon power: photon intensity: I: count per steradian per second sr −1 ⋅s −1: T −1: dn/dω: photon radiance: L q: count per square metre per steradian per second m − ...
count of photons n with energy Q p = h c/λ. [nb 2] photon flux: Φ q: count per second: s −1: T −1: photons per unit time, dn/dt with n = photon number. also called photon power: photon intensity: I: count per steradian per second sr −1 ⋅s −1: T −1: dn/dω: photon radiance: L q: count per square metre per steradian per second m − ...
Using a more accurate spectrum may give a slightly different optimum. A blackbody at 6000 K puts out 7348 W per square centimetre, so a value for u of 44% and a value of 5.73 × 10 18 photons per joule (corresponding to a band gap of 1.09 V, the value used by Shockley and Queisser) gives Q s equal to 1.85 × 10 22 photons per second per square ...
Since an incident wave of irradiance I f over an area A has a power of I f A, this implies a flux of I f /E p photons per second per unit area striking the surface. Combining this with the above expression for the momentum of a single photon, results in the same relationships between irradiance and radiation pressure described above using ...
The energy content of this volume element at 5 km from the station is 2.1 × 10 −10 × 0.109 = 2.3 × 10 −11 J, which amounts to 3.4 × 10 14 photons per (). Since 3.4 × 10 14 > 1, quantum effects do not play a role. The waves emitted by this station are well-described by the classical limit and quantum mechanics is not needed.
The number of photons per second emitted into the wedge at angle θ is I cos(θ) dΩ dA. Figure 2 represents what an observer sees. The observer directly above the area element will be seeing the scene through an aperture of area dA 0 and the area element dA will subtend a (solid) angle of d Ω 0 , which is a portion of the observer's total ...