Determining the number of photons that reach the earth

[phys62]

1. The problem statement, all variables and given/known data
Light is shining perpendicularly on the surface of the earth with an intensity of 910 W/m2. Assuming all the photons in the light have the same wavelength (in vacuum) of 668 nm, determine the number of photons per second per square meter that reach the earth.



2. Relevant equations
The power per area is:
P/A = (# of photons /t /A)*(energy / photon)

E/photon = h nu = hc / lambda

photons /t /A = (P/A) * lambda / hc


3. The attempt at a solution
photons /t /A = (P/A) * lambda / hc
photoms /t /A = (910)*(668x10^-9)/(6.63x10^-34)(3x10^-8) = 3.066x10^21

[Redbelly98]

Looks good!

Just a small typo, c is 3x10^+8. Looks like you used the correct value for the calculation though.

[phys62]

The program I enter my homework answers into doesn't let me have that many decimal places in my answer though (we have to write it out long, it doesnt let us put x10^21), so I was assuming I must've done something wrong.. :-/

[Redbelly98]

Looks like 3 significant figures are justified here, given the values of intensity and wavelength.

By the way, if you are entering numbers into a computer, you can usually write them as
6.63e-34
3e8
The "e" is pretty standard notation for computer entry, and simply means "times 10 to the ___ power"

For example:
http://www.google.com/search?hl=en&q=910*668e-9%2F(6.63e-34*3e8)&btnG=Search

[phys62]

Oh wow, I never knew I could write my answers that way on this program. haha thanks!