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

The value is
`(n)/(t) = 4.83 *10^(19) \ photons / s`

Step-by-step 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 = (E)/(t)`

=>
`P_l = (E)/(t) = (nhf)/(t)`

=>
`(n)/(t) = (P_l )/(hf)`

=>
`(n)/(t) = (16 )/(6.625 *10^(-34)* (5.00*10^(14)))`

=>
`(n)/(t) = 4.83 *10^(19) \ photons / s`