p39_011 - N = E E ph = . 93 Pt hc/ = (0 . 93)(60 W)(2 . 628...

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11. The total energy emitted by the bulb is E =0 . 93 Pt ,where P =60Wand t = 730 h = (730 h)(3600 s / h) = 2 . 628 × 10 6 s. The energy of each photon emitted is E ph = hc/λ . Therefore, the number of photons emitted is
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Unformatted text preview: N = E E ph = . 93 Pt hc/ = (0 . 93)(60 W)(2 . 628 10 6 s) (6 . 63 10 34 J s) (2 . 998 10 8 m / s) / (630 10 9 m) = 4 . 7 10 26 ....
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