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A line in the x-ray spectrum of gold has a wavelength of 18.5 pm(=18.5x10^-12 m) The emitted x-ray photons corresponds to a transition of the gold atom between two stationary states, the upper one of which has the energy -13.7 keV. What is the energy of the lower stationary state?Note) PLEASE GIVE ALL DETAILS OF SOLUTION.Note) PLEASE PAY ATTENTION: ...spectrum of gold has a wavelength of 18.5 pm(=18.5x10^-12 m)
The resolving power of a microscope depends on the wavelength used. If you wanted to "see" an atom, a wavelength of approximately 1 × 10-11 m would be required.
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