Question

An energy inefficient incandescent light bulb uses a hot filament to produce a continuous spectrum. The temperature of the filament is typically about 2853 k. At what wavelength (in nanometers)is the light bulb the brightest?

Answer 1

c

Step-by-step explanation:

Answer 2

So the wavelength of light is 1015.7 nm

Step-by-step explanation:

From Wein's displacement law, we can write the following formula;

`\lambda _(MAX)* T=K` here,
`\lambda _(MAX)` is the wavelength of maximum radiation in meters

T is the absolute temperature (in Kelvin) for maximum radiation

K is the constant has a value of
`2.898* 10^(-3)m-kelvin`

Given here, T = 2853 K

And we need to find
`\lambda _(MAX)`

Putting the values in formula

`\lambda _(max)* 2853=2.898* 10^(-3)`

`\lambda _(max)=1.0157* 10^(-6)m`

`\lambda _(max)=1015.7* 10^(-9)m=1015.7nm`