The frequency of incident photon so that the atom makes a transition from E1 to E2 should be __________ a) E2 – E1 b) E2 – E1/c c) E2 – E1/h d) E2 – E1/λ
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The frequency of incident photon so that the atom makes a transition from E1 to E2 should be __________
a) E2 – E1
b) E2 – E1/c
c) E2 – E1/h
d) E2 – E1/λ
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- An electron sitting in the -4eV energy level of this atom is struck with a photon of wavelength 345nm. As a result, the electron will -0.2 ev --0.4 ev -1.8 ev 4 ev O Absorb the photon and rise to the -0.4eV energy level Absorb the photon and rise to the -1.8eV energy level Not absorb the photon, and will remain in the -4eV energy level Absorb the photon and escape the atom Absorb the photon and rise to the -0.2eV energy levelUse the Bohr model to find the second longest wavelength of light in the Paschen series for a triply-ionized Be atom (Z = 4). Recall that the Paschen series corresponds to transitions to the second excited state (n = 3). a) 13.5 nm b) 117 nm c) 73.0 nm d) 41.1 nm e) 80.2 nmH-alpha line is a red visible spectral line in hydrogen atom with a wavelength of 656.3 nm. Consider five distant stars labeled A, B, C, D, and E. The light from these starts was detected on Earth and, after performing spectral analysis, the following H-alpha wavelengths were measured: AA = 667.5 nm, Ag = 650.4 nm, Ac = 653.5 nm, Ap = 660.3 nm, and AE = 664.9 nm. Which star has the slowest speed relative to Earth, in which direction and how fast does it move? The slowest star is? and it moves Select an answer The speed of the slowest star (in km/s), Vslowest = Which star has the fastest speed relative to Earth, in which direction and how fast does it move? The fastest star is? and it moves Select an answer Earth. The speed of the fastest star (in km/s), Vfastest Submit Question = Earth. Units Select an answer ✓ Units Select an answer ✓
- For an electron in a hydrogen atom, which of the following transitions would represent the largest quantum of energy being absorbed? Hydrogen Energy Transitions and Radiation Level n = ∞ n = 5 n = 4 486 nm n = 3 Infrared 434 nm 656 nm wavelengths n = 2 Visible wavelengths Ionization n = 1 UltravioletFind the energy of the photon released in the transition from n₁ = 6 to n₂ = 1 for a hydrogen atom. (Note: Use Rydberg Formula)If you are shining light on a Pt electrode and no electrons are emitted, what do you have to do to generate electrons? Decrease the wavelength of the light to increase the energy of the photons to emit electrons Electrons can not be emitted from Pt because like any other noble metal it does not react, even under light exposure (unlike other metals, e.g. Na, Sr or Cu) Increase the wavelength of the light to increase the energy of the photons to emit electrons Wait longer times or increase the intensity of the light source, since the energy of the absorbed light is not high enough to emit electrons from Pt surface
- The L series of the characteristic x-ray spectrum of tungsten contains wavelengths of 0.1099 nm and 0.1282 nm. The L-shell ionization energy is 11.544 keV. Which x-ray wavelength corresponds to an N → L transition? Determine the ionization energies of the M and N shells: If the incident electrons were accelerated through a 40.00 keV potential difference before striking the target, find the shortest wavelength of the emitted radiation:The Balmer series in hydrogen includes the Paschen series. has four lines in the ultraviolet. O includes both the Paschen series and the Lyman series. O has four lines in the visible. O includes the Lyman series.