Bohr devised Equation 1 to describe the emission spectrum of hydrogen:
In this equation ni is the numerical index of the level of the electron just before it emits light, while nf is the numerical index of the level where the electron winds up after the emission of light. (Note that nf does not necessarily equal 1.) DE is the difference in energy between the two energy levels in question, while RH is a proportionality constant. Since DE is related to the wavelength, l, of light by the equation DE = hc/l, Equation 1 may be written in the equivalent form of Equation 2.
In order to do the algebra required to calculate the value of l for a given transition, it is necessary to know the values of the constants RH, h, and c. They are 2.18 x 10 -18 J, 6.63 x 10-34 J sec, and 3.00 x 108 m/sec, respectively. Substituting these values into Equation 2 produces
For the transition from ni = 6 to nf = 2, DE = - 4.84 x 10-19 J. The negative sign indicates that the energy of the final state is less than that of the initial state. Converting this value of DE into a wavelength yields l = -4.10 x 10 -7 meters or 410 nm. Light of this wavelength is purple. We have ignored the negative sign since the value of l must be positive.
What type of radiation does an electron in a hydrogen atom emit when it falls from the
a. n = 6 to n = 4 level?
b. n = 2 to n = 1 level?
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