Problem 27.74 If a 120-W lightbulb emits 2.5 % of the input energy as visible li

Question

Problem 27.74

If a 120-W lightbulb emits 2.5 % 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 1.3 m away?

Express your answer using two significant figures. Answer 49.16 * 10^28 INCORRECT

Part B

How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.0 km away?

Express your answer using two significant figures. Answer 83.08*10^22 INCORRECT

Explanation / Answer

Energy emitted as visible light = 2.5% of 120 W = 3 W

Energy of 1 photon = h*c/lambda
= (6.626*10^-34 *3*10^8)/(550*10^-9)
= 3.614*10^-19 J

number of photon emitted = 3 W/ (3.614*10^-19) = 8.3*10^18 photons/ s

A)
r = 2 mm = 2*10^-3 m
number of photon hitting pupil = photon emitted * pi*r^2 / (4*pi*R^2)
= photon emitted *r^2 / (4*R^2)
= 8.3*10^18 * (2*10^-3)^2 / (4*1.3^2)
= 4.91*10^12 photon/s
Answer: 4.91*10^12 photon/s

B)
r = 2 mm = 2*10^-3 m
number of photon hitting pupil = photon emitted * pi*r^2 / (4*pi*R^2)
= photon emitted *r^2 / (4*R^2)
= 8.3*10^18 * (2*10^-3)^2 / (4*1000^2)
= 8.3*10^6 photon/s
Answer: 8.3*10^6 photon/s

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