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.
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.
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
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.
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!
oh you meant in the question, sorry they didn't copy over!
To estimate the number of photons per second of visible light that will strike the pupil of the eye, we need to calculate the rate at which energy enters the pupil and then convert that energy into the number of photons.
Let's break down the calculation step by step:
1. Find the energy of a photon with wavelength 550 nm:
- Use the energy-wavelength relationship: energy = (Planck's constant * speed of light) / wavelength.
- The Planck's constant is approximately 6.626 x 10^-34 J*s.
- The speed of light is approximately 3.00 x 10^8 m/s.
- Convert wavelength from nm to meters: 550 nm = 550 x 10^-9 m.
- Calculate the energy: energy = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (550 x 10^-9 m).
- The energy comes out to be approximately 3.62 x 10^-19 Joules.
2. Calculate the rate at which energy enters the pupil:
- Use the formula: I = P / (4πL^2), where I is the intensity, P is the power, and L is the distance.
- The power of the lightbulb is 80 W.
- The distance from the lightbulb to the observer's eye is 3.0 m.
- Calculate the intensity: I = 80 W / (4π * (3.0 m)^2).
- The intensity comes out to be approximately 2.83 W/m^2.
3. Convert the energy into the number of photons:
- Use the relationship: number of photons = energy / (Planck's constant * frequency).
- Since frequency (f) is related to energy (E) and Planck's constant (h) through the equation E = hf, the frequency can be calculated.
- The energy is 3.62 x 10^-19 J (from step 1).
- Planck's constant is approximately 6.626 x 10^-34 J*s.
- Calculate the frequency: frequency = 3.62 x 10^-19 J / (6.626 x 10^-34 J*s).
- The frequency comes out to be approximately 5.47 x 10^14 Hz.
4. Calculate the number of photons per second:
- Use the formula: number of photons per second = (intensity * effective area) / (energy per photon * frequency).
- The effective area is calculated using the formula for the area of a circle: πr^2, where r is the radius of the pupil. Given the diameter of the pupil is 4.0 mm, the radius is 2.0 mm or 2.0 x 10^-3 m.
- The energy per photon is 3.62 x 10^-19 J (from step 1).
- Plug in the values: (2.83 W/m^2 * π * (2.0 x 10^-3 m)^2) / (3.62 x 10^-19 J * 5.47 x 10^14 Hz).
- Calculate the number of photons per second: approximately 1.23 x 10^14 photons/s.
So, the estimated number of photons per second of visible light that will strike the pupil of the eye is approximately 1.23 x 10^14 photons/s.
The reason we multiply by 1.60 x 10^-19 J/eV is to convert the energy unit from joules to electron volts (eV) in order to match the energy unit of the photon's energy. This conversion factor represents the amount of energy in one electron volt.