Question

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 approximately how many visible light photons are emitted per second

Answer 1

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

   $P_l = \frac{E}{t}$

=>  $P_l = \frac{E}{t} = \frac{nhf}{t}$

=>  $\frac{n}{t} = \frac{P_l }{hf}$

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

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