1. E=h·f = h·c/λ
2. E_total = x·h·f = x·h·c/λ
       E_total corresponds to 100 W/s
       x be The Number of The Particles In the case of 500 Nm
       x = E_total · λ / (h·c)
       x = 100J·500·10^-9 m / (6.026·10^-34 JJ · 3.0·10⁸m/s)
       x = 2.5·10²⁰, d.h. about. 1/1000 mol

1. · General The idea: The of which: Energy must at least: The excitation energy corresponding to
2. · E_kin = ½ · m·v²
3.     = ½ · 9.1·10^-31·(1.5·10¹¹m/(30·3600))^2 = 8.8·10^-19 J
4. · E_A = h·c/λ = 6.626·10^-34Js·3.0·10⁸/557.7·10^-9 = 3.6·10^-19 J
5. · E_kin / E_A = about. 2.4
6.     d.h. It 's is more of which: Energy as excitation energy of which:

Estimates (Computable) from, How many photons One Light bulb with 100 W for each Seconds . In the visible Area of the Lighting emitted.
Notes 1: Take it . as Simplification to, that λ = 500 Nm

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a) E = h·c/λ = 6.626·10^-34 JJ · 299'792'458 m/s / (450·10^-9 m) = 4.4114·10^-19 J
b) E = n·h·c/λ, n = E·λ/(h·c) = 300J/s · 450·10^-9 m / (6.626·10^-34 JJ · 299792458 m/s) = 6.8·10²⁰

Electrons of the Solar winds I'll get it. after One of them Duration of flight of something . 30 h to the Earth, Where They Polar lights to call for can.
Prevailing In the case of Polar lights are green Lighting phenomena, The by: Incentives of the Oxygen In the case of λ = 557.7 Nm caused by: will be.
Wisdom Computable after, that The of which: Energy The Electrons of the Solar winds is sufficient, The Oxygen to stimulate.
Notes: The Distance Earth and Sun of which: 150 The Commission's proposal km
Notes 2: m(e^–)= 9.1·10^-31 kg

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a) Calculate The Energy One of them Photons from blue Lighting with One of them Wavelength of 450 Nm.
b) Calculate The Number of The Particles of light (photons) One of them blue Lamps with One of them Performance of 300 Watt
Watt corresponds to Joule/Seconds .

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