Question

Only a fraction of the electric energy supplied to a tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared radiation (i.e., heat). A 25−W lightbulb converts 15.0 percent of the energy supplied to it into visible light (assume the wavelength to be 505 nm). How many photons are emitted by the light bulb per second (1 W = 1 J / s)

Answer 1

The value is
`N = 9.52 *10^(18) \ photons`

Step-by-step explanation:

From the question we are told that

The power rating of the bulb is P = 25 W

The efficiency is
`\eta = 15\%`

The wavelength is
`\lambda = 505 \ nm = 505 *10^(-9) \ m`

Generally the energy of the photon emitted is mathematically represented as

`E = (h * c )/(\lambda )`

Here h is the planks constant with a value
`h = 6.63*10^(-34) \ J \cdot s`

c is the speed of light with value
`c = 3.0*10^(8) \ m/s`

So

`E = (h * c )/(\lambda )`

=>
`E = ( 6.63*10^(-34) * 3.0 *10^(8) )/(505 *10^(-9))`

=>
`E = 3.9386 *10^(-19) \ J`

Generally we are told that 1 W = 1 J/s

Hence 25 W = 25 J

Generally the amount of this energy converted to light is

`E_l = 25 * 0.15`

`E_l = 25 * 0.15`

`E_l = 3.75 \ J`

Generally the number of photons that are emitted by the light bulb per second is mathematically represented as

`N = (3.75)/( 3.9386 *10^(-19))`

=>
`N = 9.52 *10^(18) \ photons`