A light bulb of 200W emits 1.5μm. How many photons are emmited per second.

Step-by-step explanation:

An incandescent light bulb filament is approximated by a black body radiator, and in the case of a 60W bulb the filament temperature is around 2500˚C which is 2870˚K.

There are a couple of standard black body results that we can use. Firstly the total irradiance emitted per unit area of black body is equal to:

Ed=σT4

where σ=5.67×10−8 Wm−2K−4 is the Stephan-Boltzmann constant. For our bulb we get:

Ed=5.67×10−8⋅28704=3.85×106 Wm−2

As this is a 60W bulb then it has a total irradiance of 60W. Therefore the equivalent black body surface area is:

603.85×106=1.56×10−5 m2

which is 15.6mm2 of filament area.

Secondly we have that the total number of photons emitted per second per unit area of black body is equal to:[1]

Qd=σQT3

where σQ=1.52×1015 photons.sec−1m−2K−3. For our bulb this is:

Qd=1.52×1015⋅28703=3.59×1025 photons.sec−1m−2

Multiplying by the bulb’s equivalent black body surface area gives the result we require:

3.59×1025⋅1.56×10−5=5.6×1020 photons/sec

As a sanity check we know that these photons have a total energy of 60 joules per second, so the average energy per photon is:

605.6×1020=1.1×10−19 joules

A photon of wavelength λ has energy E=hcλ, and so the average energy corresponds with a photon of wavelength:

λ=hc1.1×10−19=1.9μm

Here’s a chart of the power distribution by wavelength, with the average photon wavelength shown as the dashed line, and visible wavelengths highlighted:

emember that there are a higher proportion of photons for the longer, lower energy wavelengths, so the average is weighted to the right.

We can also see from the original calculation that the general case is:

QdWEd=σQT3WσT4=2.68×1022WT photons/sec

for a bulb of wattage W watts and filament temperature T ˚K. So the photon emission rate is inversely proportional to the filament temperature. As a somewhat counter-intuitive example, a 60W halogen bulb with 3200˚K filament only emits photons at 90% of the rate of the standard 60W bulb, despite being visibly brighter.

The reason is that as the temperature increases then an increased proportion of shorter wavelength photos are emitted and therefore the average energy per photon increases, decreasing the number emitted per second. However at the same time an increased proportion of the photons are visible rather than infra-red, making the bulb appear brighter. Here’s the power distribution chart with the 60W halogen curve added for comparison:

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!

The first answer is correct