A laser pointer used in the classroom emits light at 5650 Å, at a power of 4.00 mW. (One watt is the SI unit of power, the measu

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A laser pointer used in the classroom emits light at 5650 Å, at a power of 4.00 mW. (One watt is the SI unit of power, the measu

re of energy per unit of time. 1 W = 1 J/s). How many photons are emitted from the pointer in 115 seconds?

Chemistry

2 answers:

lorasvet [2.4K] · Answer 1 · 2026-03-20T17:56:16+03:00

n equals 1.3 x 10¹⁸ photons. First, we calculate the energy released over the specified duration: E = Pt, where, E = Energy =?, P = Power = 4 mW = 0.004 W, t = time = 115 s. Thus, E = (0.004 W)(115 s), resulting in E = 0.46 J. This energy can be expressed in terms of photons: E = nhc/λ, where, n = Number of photons =?, h = Planck's Constant = 6.626 x 10⁻³⁴ J.s, c = speed of light = 3 x 10⁸ m/s, λ = wavelength of the light = 5650 x 10⁻¹⁰ m. Thus, 0.46 J = n(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5650 x 10⁻¹⁰ m), leading to n = (0.46 J)/(3.5 x 10⁻¹⁹ J), culminating in n = 1.3 x 10¹⁸ photons.

lorasvet [2.4K] · Answer 2 · 2026-03-20T17:57:01+03:00

Approximately 1.308 x 10¹⁸ photons were emitted by the laser pointer. Given: wavelength of the photon, λ = 5650 Å = 5650 x 10⁻¹⁰ m, power emitted by the source, P = 4 mW = 4 x 10⁻³ W, time of photon emission, t = 115 s. The energy of a single photon can be found using: E = hf, f = c / λ, where c represents the speed of light = 3 x 10⁸ m/s, h denotes Planck's constant = 6.626 x 10⁻³⁴ Js. The total energy from the source (laser pointer) equals P multiplied by t: = (4 x 10⁻³ W) (115 s) = 0.46 J. If no energy is lost, emitted energy from the source must match the total energy of the photons: Thus, 1.308 x 10¹⁸ photons were emitted from the laser pointer.