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An electric bulb of 100 W converts 3% of electrical energy into light energy.If the wavelength of light emitted is 6625 Å,the number of photons emitted in 1 s is……….. (h=6.625xx10^(-34))JS |
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Answer» Solution :`LAMBDA=6625Å` `=6.625xx10^(-7)m` `c=3xx10^(8)m//s` ENERGY obtained from 100 W bulb in 1s=100 J Energy converted into light energy from 3% of electric energy =3J Now E=nhf `E=(NHC)/(lambda) ` `thereforen=(Elambda)/(hc)` `=(3xx6.625xx10^(-7))/(6.625x10^(-34)xx3xx10^(8))` `therefore n=10^(19)` |
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