Consider a single photon with a wavelength of λ, a frequency of v, and an energy of E. What is the wavelength, frequency, and energy of a pulse of light containing 100 of these photons? O 100A, 100v, and 100E O 100, 100v, and E λ, v, and 100E O 0.012, v, and 100E 0.012, 0.01v, and 0.01 E

Calculate the energy for the transition of an electron from the n = 2 level to the n = 4 level of a hydrogen atom.triangle E =  _____JoulesIs this an absorption or an emission process ?  absorption or emission

By definition UV radiation from 290 to 430 nm is photon radiation with wavelengths of 290 nm more actinic and energy than higher wavelengths (higher frequencies). UV does damage to organic molecules especially macromolecules. Gamma radiation is by definition high energy photon energy and exists down in the 0.01 to 0.001 nm range. Dosage was considered as MegaRads in the past now MegaGrays (Mrad vs MGry). The Dosage of 1.5-5 MGray is used to sterilize PPE . The biokill is 2.5 MGray. Both UV and Gamma cause free radical damage to macromolecules.  Question:  Is there a sound mathematical relationship between UV radiation damage versus Gamma damage to macromolecules? Clearly UV radiation damage is dependant on the Activation Radiation of the bond dissociation energy by ev and wavelength. This is not clear with Gamma radiation but the consequence of Gamma is faster and more energetic than UV. The free radical population is higher and more substainable by Gamma than UV radiation but the…

Hi sir Please solve question 8

Which of the following energy level transitions will absorb the shortest wavelength photon? Which one emits the highest frequency photon? A. n=1→ n=3 B. n= 5 → n= 1 C. n= 6 → n= 9 D. n= 2 → n=1

what is the energy of a photon with a frequency of 5.40*1014 s-1? (h=6.626*10-34 J.s)

Please answer both parts with small explanation or leave for someone else

How many moles of photons with a frequency of 4.54 x 10^17 Hz is needed to make up 8.00kj of energy ?

What is the energy of a photon having a wavelength of 125.24 pm? Remember Planck's constant is 6.626x10-34 J×s 9.559e+11 kJ/mol   9.559e+8 kJ/mol   9.559e+4 kJ/mol   9.559e+5 kJ/mol

To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy, ?kEk , must the electrons have in order to resolve a protein molecule that is 3.70 nm3.70 nm in diameter? Take the mass of an electron to be 9.11×10−31 kg9.11×10−31 kg .