WebJan 31, 2024 · Explanation: You know, E = hc¯ν = hc λ [Where h = Planck's constant, c = velocity of electromagnetic radiation (light) and lambda = wavelength.] So, just substitute every needed value. So, E = (6.626 ⋅ 10−34) ⋅ (3 ⋅ 108) 154 ⋅ 10−9 Joules = 1.29077 ⋅ 10−18 Joules. Hence Explained. Answer link WebAnswer (1 of 5): We need to make a distinction between the energy of light and the energy of a photon. Either light of the same intensity would deliver the same energy over the …
How do you calculate the energy of a photon of ... - Socratic
WebApr 13, 2024 · d, MPP tracking of an encapsulated CF3-PEA device for over 2,000 h in air under simulated 1-sun solar illumination (equivalent to AM1.5 G, 100 mW cm −2, multi-colour light-emitting diode (LED ... WebMay 4, 2015 · Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in eV ( 1 eV = 1.602 × 10 − 19 J)? Expert Solution & Answer Want to see the full answer? Check out a sample textbook solution See solution chevron_left Previous Chapter 6, Problem 4E chevron_right Next thetford england images
8.1 Electromagnetic Energy – CHEM 1114 – Introduction to
WebApr 5, 2015 · The answer is 3.54 ×10−19 J. The equation for determining the energy of a photon of electromagnetic radiation is E = hν, where E is energy in Joules, h is Planck's … WebJul 11, 2014 · You use either the formula E = hf or E = hc λ. Explanation: h is Planck's Constant, f is the frequency, c is the speed of light, and λ is the wavelength of the radiation. EXAMPLE 1 Calculate the energy of a photon of radiation whose frequency is 5.00× 1014Hz. Solution 1 E = hf = 6.626 ×10−34J ⋅ s ×5.00 × 1014 s−1 = 3.31 × 10−19J thetford enigma soh46915