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1 Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part An electron in the hydrogen atom jumps from an orbit in which the energy is 2.9 eV higher than the energy of the final lower-energy orbit. Take Plancks constant h as 414 x 10-15 eV-s and the velocity of light as 3 x 108 ms. Calculate the frequency of the photon emitted in the transition. (You must provide an answer before moving to the next part.) (Round the final answer to four decimal places.) The frequency of the photon emitted in the transition isx 1014 Hz.
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Answer #1

frequency of the photon emiitted in the transition is

E = h*f

f = E / h

= 2.9 eV / (4.14*10^-15 eV-s )

f = 7.0*10^14 HZ

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