determine the wavelength of the second balmer line

Direct link to Just Keith's post The electron can only hav, Posted 8 years ago. We can convert the answer in part A to cm-1. So, if you passed a current through a tube containing hydrogen gas, the electrons in the hydrogen atoms are going to absorb energy and jump up to a higher energy level. The Balmer series is characterized by the electron transitioning from n3 to n=2, where n refers to the radial quantum number or principal quantum number of the electron. Number a. Direct link to BrownKev787's post In a hydrogen atom, why w, Posted 8 years ago. For this transition, the n values for the upper and lower levels are 4 and 2, respectively. And if an electron fell Express your answer to two significant figures and include the appropriate units. this Balmer Rydberg equation using the Bohr equation, using the Bohr model, I should say, the Bohr model is what So that's eight two two It was also found that excited electrons from shells with n greater than 6 could jump to the n=2 shell, emitting shades of ultraviolet when doing so. The values for \(n_2\) and wavenumber \(\widetilde{\nu}\) for this series would be: Do you know in what region of the electromagnetic radiation these lines are? However, all solids and liquids at room temperature emit and absorb long-wavelength radiation in the infrared (IR) range of the electromagnetic spectrum, and the spectra are continuous, described accurately by the formula for the Planck black body spectrum. The wavelength of the first line of Balmer series is 6563 . (n=4 to n=2 transition) using the For example, the (\(n_1=1/n_2=2\)) line is called "Lyman-alpha" (Ly-), while the (\(n_1=3/n_2=7\)) line is called "Paschen-delta" (Pa-). So I call this equation the Experts are tested by Chegg as specialists in their subject area. For example, the tungsten filaments in incandescent light bulbs give off all colours of the visible spectrum (although most of the electrical energy ends up emitted as infrared (IR) photons, explaining why tungsten filament light bulbs are only 5-10% energy efficient). Determine likewise the wavelength of the third Lyman line. B This wavelength is in the ultraviolet region of the spectrum. So to solve for lamda, all we need to do is take one over that number. Locate the region of the electromagnetic spectrum corresponding to the calculated wavelength. For an electron to jump from one energy level to another it needs the exact amount of energy. In which region of the spectrum does it lie? Direct link to ANTHNO67's post My textbook says that the, Posted 8 years ago. 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