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 75-W lightbulb converts 15.0 percent of the energy supplied to it into visible light (assume the wavelength to be 550 nm). How many photons are emitted by the lightbulb per second (1 W = 1 J/s)?

Answer 1

`2.8317205*10^(+18)` photons

Step-by-step explanation:

Infrared radiation also consists of photons, but we will assume they mean visible photons.

The energy of a photon is given by the formula
`E_p=hf`, where f is the frequency of the photon and
`h=6.626*10^(-34)m^2kg/s` is Planck's constant.

The relationship between frequency and wavelength is given by
`f=(c)/(\lambda)`, where c=299792458m/s is the speed of light.

The energy of N photons will be then given by:

`E_N=Nhf=(Nhc)/(\lambda)`

Which can be written as:

`N=(E_N\lambda)/(hc)`

We need to know how much energy do we have. The power we have relates to the energy by the equation P=E/t, but of this energy we only get 15% as visible, so the energy related to the visible photons will be 0.15E.

Putting all together:

`N=(E_N\lambda)/(hc)=(0.15E\lambda)/(hc)=(0.15Pt\lambda)/(hc)`

We then substitute our values, considering only 1s:

`N=(0.15(75W)(1s)(550*10^(-9)m))/((6.626*10^(-34)m^2kg/s)(299792458m/s))=2.8317205*10^(+18)`