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Chemistry Question on Structure of atom

A 150 watt bulb emits light of wavelength 6600 ?? and only 8% of the energy is emitted as light. How many photons are emitted by the bulb per second?

A

4×10194 \times 10^{19}

B

3.24×10193.24 \times 10^{19}

C

4.23×10204.23 \times 10^{20}

D

3×10203 \times 10^{20}

Answer

4×10194 \times 10^{19}

Explanation

Solution

Power of the bulb = 150 watt=150 J s−1 {150\, watt = 150 \, J\, s^{-1}}
As only 8% of the energy is emitted as light so, the total energy emitted per second.
=150 J×8100=12 J= \frac{150 \, J \times 8}{100} = 12 \, J
Energy of one photon, E=hv=hcλE = hv = \frac{hc}{\lambda}
=(6.626×10−34Js)×(3×108ms−1)6600×10−10m= \frac{\left(6.626 \times10^{-34} J s\right) \times \left(3\times 10^{8} m s^{-1}\right)}{6600 \times 10^{-10} m}
=3.0118×10−19J= 3.0118 \times 10^{-19} J
∴\therefore Number of photons emitted
=12J3.0118×10−19J=3.98×1019≈4.0×1019= \frac{12J}{3.0118 \times 10^{-19} J} = 3.98 \times 10^{19} \approx4.0 \times 10^{19 }