Constants: Planck’s constant: h = 6.626×10^-34 J-s Speed of light: c = 3.00×10^8 m/s nm = 10^-9m Equations: V = c/λ Eph = hv = hc/λ AE = E – Ei = 22.179×10^-18 J (1/n^2) Calculate the energy of a photon that has a wavelength of 732 nm.
The Correct Answer and Explanation is:
To calculate the energy of a photon with a wavelength of 732 nm, we use the equation:Eph=hcλE_{ph} = \frac{hc}{\lambda}Eph=λhc
Step 1: Convert the wavelength to meters
λ=732 nm=732×10−9 m=7.32×10−7 m\lambda = 732 \, \text{nm} = 732 \times 10^{-9} \, \text{m} = 7.32 \times 10^{-7} \, \text{m}λ=732nm=732×10−9m=7.32×10−7m
Step 2: Plug in the known values
- Planck’s constant h=6.626×10−34 J\cdotpsh = 6.626 \times 10^{-34} \, \text{J·s}h=6.626×10−34J\cdotps
- Speed of light c=3.00×108 m/sc = 3.00 \times 10^8 \, \text{m/s}c=3.00×108m/s
Eph=(6.626×10−34)(3.00×108)7.32×10−7E_{ph} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{7.32 \times 10^{-7}}Eph=7.32×10−7(6.626×10−34)(3.00×108)
Step 3: Calculate the numerator
6.626×10−34×3.00×108=1.9878×10−256.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25}6.626×10−34×3.00×108=1.9878×10−25
Step 4: Divide by the wavelength
Eph=1.9878×10−257.32×10−7≈2.715×10−19 JE_{ph} = \frac{1.9878 \times 10^{-25}}{7.32 \times 10^{-7}} \approx 2.715 \times 10^{-19} \, \text{J}Eph=7.32×10−71.9878×10−25≈2.715×10−19J
Final Answer:
2.72×10−19 J\boxed{2.72 \times 10^{-19} \, \text{J}}2.72×10−19J
Explanation
Photons are particles of light that carry energy, and this energy depends directly on the frequency or inversely on the wavelength of the light. The fundamental equation used to calculate the energy of a photon is:E=hcλE = \frac{hc}{\lambda}E=λhc
Where hhh is Planck’s constant and ccc is the speed of light. The wavelength (λ\lambdaλ) must always be in meters to be consistent with the SI units of hhh and ccc.
In this problem, we are given a wavelength of 732 nanometers. Since 1 nanometer equals 10−910^{-9}10−9 meters, we convert 732 nm to 7.32×10−77.32 \times 10^{-7}7.32×10−7 meters. Plugging this value into the equation along with the known constants yields a photon energy of approximately 2.72×10−192.72 \times 10^{-19}2.72×10−19 joules.
Among the answer choices, 2.72e-19 J is the only one that matches this calculation. The other options are significantly off and can be ruled out by checking the order of magnitude alone.
This calculation is fundamental in quantum physics and is often used in spectroscopy, astronomy, and understanding atomic transitions. It helps scientists determine how much energy is involved when photons are emitted or absorbed by atoms. In practical terms, it connects the microscopic world of quantum particles to observable phenomena like light color and photoelectric effects.
