Answer to Question #234774 in Electricity and Magnetism for Sridhar

Question

If only 8% of the energy supplied to a bulb is radiated as visible light then number of photons emitted per second by a 100 watt bulb are (Assume wavelength of visible light is 5.6×10^(-7)m )

Answer 1

$λ=5.6 \times 10^{-7} \;m \\ h=6.625 \times 10^{-34} \;m^2gr/s$

Energy of one photone $= \frac{hc}{λ}$

$= \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{5.6 \times 10^{-10}} \\ = 3.549 \times 10^{-19} \;J$

A 100 Watt bulb has to be supplied 100 J of energy.

Energy radiated by bulb per second as visible light $= \frac{100 \times 8}{100}= 8 \;J$

Visible light photons emitted per second as visible light $= \frac{8}{3.549 \times 10^{-19}}$

$=2.254 \times 10^{19 }$