p41_070 - 70. (a) The energy difference between the two...

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Unformatted text preview: 70. (a) The energy difference between the two states 1 and 2 was equal to the energy of the photon emitted. Since the photon frequency was f = 1666 MHz, its energy was given by hf = (4.14 × 10−15 eV · s)(1666 MHz) = 6.90 × 10−6 eV. Thus, E2 − E1 = hf = 6.9 × 10−6 eV = 6.9 µeV . (b) The emission was in the radio region of the electromagnetic spectrum. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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