If a 80 lightbulb emits 4.0 of the input energy as visible light (average wavelength 550 ) uniformly in all directions, estimate how many photons per second of visible light will strike the pupil (4.0 diameter) of the eye of an observer 3.0 away.

Question

1. I found the energy of photon 1.24*10^3 eV*nm/ 550 nm = 2.25

2. i know I = P/4piL^2 so the rate energy enters pupil is I pi r^2 which equals Pr^2/4L^2 so the rate photons enter should be n/t = (E/t)/hf or Pr^2/4L^2hf

3. (0.04)(80 W)(4.0 * 10^-3 m) / 4(3.0 *10^-3 m)^2(2.25 eV)(1.60 * 10^-19 J/eV)

which gives me 3.95 * 10^6 photons/s which is not the correct answer according to mastering physics.

I do not particularly see why 1.60 * 10^-19 should be there but someone told me I had to put it there.

Answer 1

That 2.25eV has to be converted to Joules

I wish you had included units on the numbers, so I could understand your work.

numberphotons*energy/photon= 80watts*(diameter/distance)^2

I just don't see diameter squared in there, but without units, I don't know

Answer 2

I believe that P(total) =80 W, P = 0.04•P(total) = 0.04•80 = 3.2 W,
the wavelength λ = 550 nm, d = 4 mm, L = 3 (? units), h =6.62•10^-34 J•s is the Planck’s constant, c = 3•10^8 m/s is the speed of light.
Then
0.04•P = E/t = n•E(o)/t = n•h•c/ λ•t.
n = 0.04•P• λ•t/h•c,
This number of photons will strike the sphere of radius L (4πL^2), therefore, the number of photons that will strike the area π•d^2/4 of this sphere is
n(o) = 0.04•P• λ•t• π•d^2/ h•c•16•π•L^2.
The number of photons per second is
n(o)/t = 0.04•P• λ•d^2/ h•c•16• L^2.

Answer 3

what units were missing? I see units on all my numbers. I found it though, I just had to divide my radius in half... although I do not know why!

Answer 4

oh you meant in the question, sorry they didn't copy over!