When UV radiation of wavelength 58.4 nm from a helium lamp is directed on to a sample of Xe, electrons are ejected with a speed of 1.79 × 106 m s−1. Calculate the ionisation energy of Xe.
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When UV radiation of wavelength 58.4 nm from a helium lamp is directed on to a sample of Xe, electrons are ejected with a speed of 1.79 × 106 m s−1. Calculate the ionisation energy of Xe.
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- What is the electron configuration of the Ba3+ ion? Suggest a reason why this ion is not normally found in nature.Palladium, with an electron configuration of [Kr] 4d10, is an exception to the aufbau principle. Write the electron configuration of the 2+ cation of palladium. Does the fact that palladium is an exception influence the electron configuration of Pd2+?Photoelectron spectroscopy applies the principle of the pho-toelectric effect to study orbital energies of atoms and mol-ecules. High-energy radiation (usually UV or x-ray) is absorbedby a sample and an electron is ejected. The orbital energy can becalculated from the known energy of the radiation and the mea-sured energy of the electron lost. The following energy differ-ences were determined for several electron transitions:ΔE 2→1=4.098X10⁻¹⁷J, ΔE 3→1=4.854X10⁻¹⁷J, ΔE 5→1=5.242X10⁻¹⁷J, ΔE 4→2=1.024X10⁻¹⁷J Calculate the energy change and the wavelength of a photon emitted in the following transitions:(a) Level 3→2 (b) Level4→1 (c) Level5→4
- How much energy is required to excite an electron from a n3 ⟶⟶ n5 within the hydrogen atom?What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.90×1015Hz? Express your answer in joules to three significant figures. It is not 9.17×10−19 J or 12.589 x 10-19 J By using the following formula, you can calculate the kinetic energy of the emitted electron: KE=E−ϕ=hν−hν0 where h=6.63×10−34 J⋅s is Planck's constant, ν=1.90×1015Hz is the given frequency, and ν0=9.39×1014 Hz is your answer from Part A. The threshold frequency ν0 of cesium is 9.39×1014 HzIn a future hydrogen-fuel economy, the cheapest source of H₂ will certainly be water. It takes 467 kJ to produce 1 mol of Hatoms from water. What is the frequency, wavelength, and mini-mum energy of a photon that can free an H atom from water?
- It takes 163. kJ/mol to break an nitrogen-nitrogen single bond. Calculate the maximum wavelength of light for which an nitrogen-nitrogen single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. nm x10 GThe energy required to dissociate the H2 molecule to H atoms is 432 kJ/mol H2. If the dissociation of an H2 molecule were accomplished by the absorption of a single photon whose energy was exactly the quantity required, what would be its wavelength (in meters)?The ionization energy of potassium is 4.34 eV and the electron affinity of Br is 3.36 eV. What is the change in energy in the reaction K(g) + Br(g) → K+(g) + Bl-(g)?
- In photoelectron spectroscopy (PES), ultraviolet radiation is directed at an atom or a molecule. Electrons are ejected from the valence shell and their kinetic energies are measured. Because the energy of the incoming ultraviolet photon is known and the kinetic energy of the outgoing electron is measured, the ionization energy, I, can be deduced from the fact that the total energy is conserved. The speed of the ejected electron, v, and the frequency of the incoming radiation, v, are related by the equation hv = I + m,u? where h is Planck's constant and m, is the mass of an electron. Show that this relationship is true by completing the statement. The energy of the incoming photon, Eotal, is conserved in photoelectron spectroscopy. Therefore, the energy of the incoming photon is equal to the plus the , so The energy of the incoming photon can also be expressed as and the kinetic energy of the outgoing electron can be expressed as . The energy required to eject the electron corresponds…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.The ionization energy of lithium is 520.2 kJ/mole, and the electron affinity of hydrogen is 72.8 kJ/mole.(a) Find the separation distance in LiH at which the Coulomb potential energy equals the energy cost of removing an electron from Li and adding it to H.(b) The measured electric dipole moment of the molecule LiH is 2.00 × 10^−29 C · m. What is the fractional ionic character of LiH?(c) Instead of removing an electron from Li and attaching it to H, we could regard the formation of LiH as occurring by removing an electron from H and attaching it to Li (electron affinity = 59.6 kJ/mole). Why don’t we consider this as the formation process?