Question

A bulb emits light of wavelength $\lambda =4500A˚$. The bulb is rated as $150watt$ and $8%$ of the energy is emitted as light. How many photons are emitted by the bulb per second?

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Solution

Step 1: Given dataThe wavelength of the light is $\lambda =4500A˚$.The power of the light is $P=150watt.$The energy emitted by the light is $8%$.Step 2: Concept and formulaeThe energy of a photon of wavelength $\lambda$ is $E=\frac{hc}{\lambda }$, where, h is the Plank constant and c is the velocity of light in free space.Power is the total amount of energy emitted in a unit of time, $P=\frac{E}{t}$.Step 3: Finding the number of photonsLet n number of photons being emitted per second by the bulb.Considering $3×{10}^{8}m/s$ is the velocity of light in free space and $6.626×{10}^{-34}j.s$ is the Plank constant.The energy emitted by the bulb is, $∆E=150×\frac{8}{100}=12\phantom{\rule{0ex}{0ex}}∆E=12joule.$The energy of 1 photon is,$E=\frac{hc}{\lambda }=\frac{\left(6.626×{10}^{-34}\right)×3×{10}^{8}}{4500×{10}^{-10}}=4.42×{10}^{-19}joule.\phantom{\rule{0ex}{0ex}}orE=4.42×{10}^{-19}joule.$According to question light(photon) energy = 12 joule,$nE=12joule\phantom{\rule{0ex}{0ex}}orn×4.42×{10}^{19}=12.\phantom{\rule{0ex}{0ex}}orn=\frac{12}{4.42×{10}^{-19}}=2.72×{10}^{19}.\phantom{\rule{0ex}{0ex}}orn=2.72×{10}^{19}.$Therefore, the number of photons emitted by the bulb is $n=2.72×{10}^{19}.$

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