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nirvana33 [79]
3 years ago
9

If the average frequency emitted by a 160 W light bulb is 5.00 1014Hz and 10.0 of the input power is emitted as visible light ap

proximately how many visible light photons are emitted per second
Physics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

The value is \frac{n}{t}  = 4.83 *10^{19} \  photons / s

Explanation:

From the question we are told that

   The power rating of the bulb is  P = 160 \ W

   The frequency is f =  5.00 *10^{14} \ Hz

   The percentage of the input power that is emitted as visible light is \eta =  10\% = 0.10

   

Generally the amount of power emitted as visible light is mathematically represented as

       P_l =  0.10 * P_i

=>  P_l =  0.10 *160

=>  P_l =  16 \ W

Generally the amount of energy emitted as light is mathematically represented as

        E = n *  h  *  f

Here n is the number of photon ,  h is the Planks constant with value h =  6.625*10^{-34} \  J\cdot s

Generally this power emitted as visible light is mathematically represented as

   P_l = \frac{E}{t}

=>  P_l = \frac{E}{t} = \frac{nhf}{t}

=>  \frac{n}{t}  = \frac{P_l }{hf}

=>  \frac{n}{t}  = \frac{16 }{6.625 *10^{-34}* (5.00*10^{14})}

=>  \frac{n}{t}  = 4.83 *10^{19} \  photons / s

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