Question

KCh 27 HW Item 17 17 of 28> ns If a 100-W lightbulb emits 3.0 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions Part A How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 2.5 m away? Express your answer using two significant figures A6.629 10 photons/s Submit Previous Answers Request Answer XIncorrect; Try Again; 5 attempts remaining Part B How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.3 km away? Express your answer using two significant figures photons/s Submit Request Answen Provide Feedback Next >

Answer 1

Solution)

A)

Power = work/time

Power= 100J/s *3/100% => 3J/s

E = hf = hc/lambda

E = 3.6142*10^(-19)J/photon. So, we divide and find that there are going to be 8.3*10^(18)photons/s to be emitted.
Now we want to find the fractional part of the surface area that the eye occupies.
pi*(0.002m)^2 = area of eye

4pi*(2.5m)^2 = total area
(0.002)^2/4(2.5)^2 * 8.3*10^(18) => 1.927*10^(-26) photons/s. (Ans)

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B) 1.3 km away

pi*(0.002m)^2 = area of eye

4pi*(1300 m)^2 = total area
(0.002)^2/4(1300)^2 * 8.3*10^(18)

=>7.12*10^(-32) photons/s. (Ans)

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Good luck!:)