Calculate how much energy is contained in a pulse of light that contains 4560 photons at a wavelength of 650 nm.

Question

BlackTom AI · Accepted Answer

First, recall that the energy of a single photon is given by E = h c / λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength.
Option 1: Compute energy per photon for λ = 650 nm. Using h ≈ 6.626×10^-34 J·s and c ≈ 3.00×10^8 m/s, E_per_photon ≈ (6.626×10^-34)(3.00×10^8) / (650×10^-9) ≈ 3.06×10^-19 J.
Now multiply by the number of photons: 3.06×10^-19 J × 4560 ≈ 1.40×10^-15 J. This matches the magnitude and value of 1.4×10^-15 J.

Option 2: If one tries 2.0×10^-27 J, compare to the per-photon energy: this is far too small; a single photon at visible wavelengths is on the order of 10^-19 J, so 10^-27 J would correspond to an almost negligible energy, clearly inconsistent with 4560 photons at 650 nm.

Option 3: 1.4×10^-33 J is even more orders of magnitude smaller than the correct total energy. Since even a single photon at 650 nm is ~3×10^-19 J, multiplying by 4560 would still be around 10^-15 J, not 10^-33 J.

Option 4: 'None of these' would require that none of the listed values match the calculation. Given that 1.4×10^-15 J precisely aligns with the computed result, this option is not correct.

Option 5: 1.5×10^-22 J lies between the per-photon energy estimate and the total energy, but it does not equal the product for 4560 photons at 650 nm; it undervalues the total energy by several orders of magnitude.

In summary, evaluating the energy per photon and scaling by the photon count leads to approximately 1.4×10^-15 J, which is consistent with the first option and inconsistent with the others.

Calculate how much energy is contained in a pulse of light that contains 4560 photons at a wavelength of 650 nm.单项选择题

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