Argon Beam

Chemistry question (5 stars!)?

An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532nm. The diameter of the laser beam is 5.5mm if the laser is pointed toward a pinhole with a diameter of 1.2mm how many photons will travel through the pinhole per second? assume that the light intensity is equally distributed throughout the entire cross-sectional area of the beam. (1 W = 1 J/s)

I solved for the energy of the photon to then calculate the number of photons per second of the entire beam: 1.33 x 10^20 photons/sec

but i cant seem to calculate the number of photons going through the pinhole, i calculated the areas of both circles and got a close answer of 6.38 x 10^18 but the answer is 6.4 x 10^17

what did i do wrong? how do i do this help 5 stars

You made a mistake in calculating the number of photons/sec in the 5.0 W laser beam. The answer is 1.33 × 10¹⁹ - you were too high by a factor of 10. That explains why your final answer is a factor of 10 too high.

Without cranking through the formulas and conversion factors (which I use all the time and are in my head and you clearly know how to use already, since you got as close as you did), here's how it goes:

λ = 532 nm ⇒ 1/λ = 18,796 cm⁻¹ ⇒ E = hc/λ = 2.33 eV/photon ⇒ 224856 J/mol of photons

∴ 5 W = 5 J/s ~ (5 J/s)/(224856 J/mol) = 2.223 × 10⁻⁵ mol of photons = 1.33 × 10¹⁹ photons

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laser pinhole problem?

If the laser is pointed toward a pinhole with a diameter of 1.1 mm, how many photons will travel through the pinhole per second? Assume that the light intensity is equally distributed throughout the entire cross-sectional area of the beam. (1 W= 1 J/s)

An argon ion laser puts out 5.3 W of continuous power at a wavelength of 532nm . The diameter of the laser beam is 6.3mm .

It's been a while:

E= h x c / lambda

Energy of a Photon = (plank's constant)(speed of light)/wavelength in meters

Assuming:
h = 6.63E-34 Joule Seconds
c = 3.00E8 m/s
wavelength = 532E-9m or 5.32E-7 m

Find Energy of a Photon @ a wavelength of 532nm:
E = [(6.63E-34) Js * (3.00E8) m/s] / (5.32E-7) m
= [ 1.989E-25] J m / 5.32E-7m
= 3.74E-19 Joules per photon

Power is given as 5.3W = 5.3 Joules / Second.

Now we can figure out how many photons pass through a cross section of the original beam:

(5.3 J / s) / (3.74E-19 Joules / photon)
= 1.42E19 photons per second.

Photon density can be assumed to be uniform, as it's given that they're evenly distributed through the original beam. Assuming that wave interference doesn't play into it, the photon density of the masked beam should be the same, as the mask's aperture is smaller than the original beam diameter. The number of photons passing through per second is going to be directly proportional to the square of the radius.

so,

1.42E19 (photons/s) / 6.3^2 (m^2)= x / 1.1^2 (m^2)

Solve for X. As my old physics and computer science profs loved to say, "the rest left as an exercise for the student"

I suggest you check the math and the logic if you intend to submit this for grades.