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Physics

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Answer

3.61×10−19 J3.61 \times 10^{-19} \text{ J}

Step-by-step solution

4 steps

Here is a quantum question:

Calculate the energy of a photon of green light with a wavelength of 550 nm550 \text{ nm}.

  • Planck's constant (hh) = 6.626×10−34 J⋅s6.626 \times 10^{-34} \text{ J}\cdot\text{s}
  • Speed of light (cc) = 3.00×108 m/s3.00 \times 10^8 \text{ m/s}
  1. Identify the given values and constants.

    • Wavelength (λ\lambda) = 550 nm550 \text{ nm}
    • Planck's constant (hh) = 6.626×10−34 J⋅s6.626 \times 10^{-34} \text{ J}\cdot\text{s}
    • Speed of light (cc) = 3.00×108 m/s3.00 \times 10^8 \text{ m/s}
  2. Convert the wavelength from nanometers to meters.

    1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m} λ=550 nm×10−9 m1 nm\lambda = 550 \text{ nm} \times \frac{10^{-9} \text{ m}}{1 \text{ nm}} λ=5.50×10−7 m\lambda = 5.50 \times 10^{-7} \text{ m}
  3. Calculate the frequency (ff) of the photon using the relationship between speed of light, wavelength, and frequency.

    The formula is c=λfc = \lambda f.
    Rearrange to solve for ff:

    f=cλf = \frac{c}{\lambda}

    Substitute the values:

    f=3.00×108 m/s5.50×10−7 mf = \frac{3.00 \times 10^8 \text{ m/s}}{5.50 \times 10^{-7} \text{ m}} f≈5.4545×1014 Hzf \approx 5.4545 \times 10^{14} \text{ Hz}
  4. Calculate the energy (EE) of the photon using Planck's equation.

    The formula is E=hfE = hf.
    Substitute the values:

    E=(6.626×10−34 J⋅s)×(5.4545×1014 Hz)E = (6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (5.4545 \times 10^{14} \text{ Hz}) E≈3.61×10−19 JE \approx 3.61 \times 10^{-19} \text{ J}

    The energy of the photon is 3.61×10−19 J3.61 \times 10^{-19} \text{ J}.

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