Calculate the de Broglie wavelength associated with the electron in the 2nd excited state of hydrogen atom. The ground state energy of the hydrogen atom is 13.6 eV.
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- An electron in a Hydrogen atom is in energy level n=5 and drops down into the n=2 level. What is the wavelength of the photon that is emitted in units of nm.The electron in a hydrogen atom falls from an excited energy level to the ground state in two steps, causing the emission of photons with wavelengths of 93.79 nm and 7503.1 nm, respectively. What is the principal quantum number of the initial excited energy level from which the electron falls?Consider photons incident on a hydrogen atom. (a) A transition from the n = 4 to the n = 7 excited-state requires the absorption of a photon of what minimum energy? eV(b) A transition from the n = 1 ground state to the n = 6 excited state requires the absorption of a photon of what minimum energy? eV
- The wavelength of a red spectral emission line is 670.0 nm. At what kinetic energy (in J) would an electron have that wavelength as its de Broglie wavelength?Hydrogen gas can be placed inside a strong magnetic field B=12T. The energy of 1s electron in hydrogen atom is 13.6 eV ( 1eV= 1.6*10 J ). a) What is a wavelength of radiation corresponding to a transition between 2p and 1s levels when magnetic field is zero? b) What is a magnetic moment of the atom with its electron initially in s state and in p state? c) What is the wavelength change for the transition from p- to s- if magnetic field is turned on?The transition from the first excited state to the ground state in potassium results in the emission of a photon with lambda= 770 nm. If potassium vapor is used in a Franck-Hertz experiment, at what voltage would you expect to see the first decrease in current?