A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate

the number of photons emitted per second by the bulb.

Given,
   Power of the bulb = 100 watt = work/time = 100 J/s
        Energy of one proton,         E = hcλh = 6.626×10-34Jsc = 3×108m/sλ = 400×10-9mTherefore,     E= 6.626×10-34Js×3×108m/s400×10-9m                         = 4.969 × 10-19 JNo. of photos emitted per second                         = 100 J/s4.969×10-19J= 2.012×1020 s-1

12. 
E=hcλor, λ=hcEE= 1eV = 1.60×10-19 JTherefore, E=6.626×10-34×3×1081.60×10-19=12.42×10-7 m= 0.0124×10-10 m = 0.0124 A°

  • 36

2*10^20

  • -7

HERE IS UR ANSWER....HOPE IT HELPED ;)
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