A bulb emits light of wavelength nm. The bulb is rated as 200 W and 14% of the energy is emitted as — Atomic Structure Chemistry Question

💡 Solution & Explanation

Useful power emitting light $= 0.14 \times 200 = 28$ W. Number of photons per second $n = \frac{P \lambda}{hc} = \frac{28 \times (1987.8 / 7) \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = \frac{4 \times 1987.8 \times 10^{-9}}{19.878 \times 10^{-26}} = 4 \times 10^{19}$. Therefore, correct answer is D.