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A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb. |
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Answer» Solution :Power of the BULB = 100 WATT `= 100 J s^(-1)` Energy of one photon, `E = hv = (hc)/(lamda)((6.626 xx 10^(-34)Js) xx (3 xx 10^(8) ms^(-1)))/(400 xx 10^(-9)m) = 4.969 xx 10^(-19) J` `:.` NUMBER of photons emitted `= (100 Js^(-1))/(4.969 xx 10^(-19)J) = 2.012 xx 10^(20) s^(-1)` |
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