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If a monochromatic light of frequency 6.0×1014 Hz is produced by a source and the power emitted is 2.0×10-3W, then the number of photons emitted per secnod by the source is(a) 3×1010 (b) 5×1015(c) 7×1012 (d) 9×1012 |
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Answer» Correct Option (b) 5×1015 Explanation : An energy, E = hv = 6.63×10–34 x 6.05×1014 = 3.98×10–19J Now, number of photons emitted per second n = p/E = 2.0 x 10-3/3.98 x 10-19 J = 5 x 1015 photons per second
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