(a) Interpretation: To determine the energy of the photons that is necessary to bring about the given reaction. Concept introduction: Planck has described the quantization of energy using his famous equation which is also known as the Planck’s equation as shown below; E=hν Where h is also known as the Planck’s constant and is a constant of proportionality. E is the energy of the photon and ν is the frequency of the light .
(a) Interpretation: To determine the energy of the photons that is necessary to bring about the given reaction. Concept introduction: Planck has described the quantization of energy using his famous equation which is also known as the Planck’s equation as shown below; E=hν Where h is also known as the Planck’s constant and is a constant of proportionality. E is the energy of the photon and ν is the frequency of the light .
Solution Summary: The author explains Planck's quantization of energy using his famous equation.
Photoelectron spectroscopy applies the principle of the photoelectric effect to study orbital energies of atoms and mol ecules. High-energy radiation (usually UV or x-ray) is absorbed by a sample and an electron is ejected. The orbital energy can be calculated from the known energy of the radiation and the mea sured energy of the electron lost. The following energy differences were determined for several electron transitions:
∆E2→1 = 4.098x10-17 J ∆E3→£1 = 4.854x10-17 J ∆E5→1 = 5.242x10-17 J ∆E4→2 = 1.024x10-17 J Calculate ∆E and l of a photon emitted in the following transitions:
(a) level 3 → 2; (b) level 4 → 1; (c) level 5→4.
5. Energy from radiation can cause chemical bonds to break. To break a nitrogen-nitrogen bond in N2 gas, 941kJ/mole is required.
a. Calculate the wavelength of radiation that could break the bond.
b. In what spectral range does this radiation occur?
Photoelectron spectroscopy applies the principle of the photoelectric effect to study orbital energies of atoms and molecules. High-energy radiation (usually UV or X-ray) is absorbed by a sample and an electron is ejected. The orbital energy can be calculated from the known energy of the radiation and the measured energy of the electron lost. The following energy differences were determined for several electron transitions:
ΔE2 →1 = 4.098 ×10−17 J
ΔE3 →1 = 4.854 × 10−17 JΔE5 → 1 = 5.242 ×10−17 J
ΔE4 → 2 = 1.024 ×10−17 J Calculate the energy change and the wavelength of a photon emitted in the following transitions. Enter your answers in scientific notation. Use 6.626 ×10−34 J·s for Planck's constant. (a) Level 3 to 2: ______J ______m (b) Level 4 to 1: _____J_____ m (c) Level 5 to 4: _____J
_____m
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell