Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Properties of Photons - College Chemistry

  1. A photon with wavelength 589 nm is absorbed by a sodium atom. What is the energy of this photon in joules? (Planck's constant h=6.626×10−34h = 6.626 \times 10^{-34} J·s, speed of light c=3.00×108c = 3.00 \times 10^8 m/s)
    Answer: 3.37×10−193.37 \times 10^{-19} J When you encounter photon energy problems, you're working with the fundamental relationship between electromagnetic radiation's wavelength and energy. This connects quantum mechanics to spectroscopy, which is crucial for understanding atomic transitions. To find photon energy, you use the equation E=hcλE = \frac{hc}{\lambda}, where hh is Planck's constant, cc is the speed of light, and λ\lambda is wavelength. First, convert the wavelength to meters: 589 nm = 589×10−9589 \times 10^{-9} m = 5.89×10−75.89 \times 10^{-7} m. Now substitute the values: E=(6.626×10−34)(3.00×108)5.89×10−7E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.89 \times 10^{-7}} Calculate the numerator: (6.626×10−34)(3.00×108)=1.988×10−25(6.626 \times 10^{-34})(3.00 \times 10^8) = 1.988 \times 10^{-25} Divide by the wavelength: E=1.988×10−255.89×10−7=3.37×10−19E = \frac{1.988 \times 10^{-25}}{5.89 \times 10^{-7}} = 3.37 \times 10^{-19} J This confirms answer A is correct. Answer B (3.37×10−183.37 \times 10^{-18} J) is off by a factor of 10, likely from a unit conversion error. Answer C (3.37×10−203.37 \times 10^{-20} J) is too small by a factor of 10, possibly from incorrectly handling the wavelength conversion. Answer D (3.37×10−213.37 \times 10^{-21} J) is off by two orders of magnitude, suggesting multiple calculation errors. Remember: always convert wavelength to meters before calculating, and photon energies for visible light typically fall in the 10−1910^{-19} to 10−1810^{-18} J range.
  2. A photon has energy 2.48×10−182.48 \times 10^{-18} J. What is the wavelength of this photon in nanometers?
    Answer: 80.1 nm This question tests your understanding of the fundamental relationship between a photon's energy and wavelength, described by Planck's equation and the speed of light. To find the wavelength, you need to combine two key equations. First, Planck's equation: E=hfE = hf, where h=6.626×10−34h = 6.626 \times 10^{-34} J·s and ff is frequency. Second, the wave equation: c=λfc = λf, where c=3.00×108c = 3.00 \times 10^8 m/s and λλ is wavelength. Combining these gives: E=hcλE = \frac{hc}{λ}, which rearranges to λ=hcEλ = \frac{hc}{E}. Substituting the values: λ=(6.626×10−34)(3.00×108)2.48×10−18=8.01×10−8λ = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{2.48 \times 10^{-18}} = 8.01 \times 10^{-8} m. Converting to nanometers by multiplying by 10910^9: λ=80.1λ = 80.1 nm. Answer A (80.1 nm) is correct based on this calculation. Answer B (125 nm) suggests you might have used an incorrect value for one of the constants or made an arithmetic error. Answer C (248 nm) could result from confusing the given energy value with the wavelength or using incorrect unit conversions. Answer D (801 nm) indicates a decimal place error in the final conversion to nanometers. Study tip: Memorize the combined formula λ=hcEλ = \frac{hc}{E} for photon problems. Always double-check your unit conversions—wavelength problems frequently involve switching between meters, nanometers, and other units. Practice the arithmetic with scientific notation to avoid calculation errors.
  3. A photon detector measures photons with energies ranging from 1.0×10−191.0 \times 10^{-19} J to 5.0×10−195.0 \times 10^{-19} J. What is the corresponding frequency range that this detector can measure?
    Answer: 1.5×10141.5 \times 10^{14} Hz to 7.5×10147.5 \times 10^{14} Hz When you encounter questions about photon energy and frequency, you're dealing with one of the fundamental relationships in quantum mechanics. The key equation connecting these properties is Planck's equation: E=hfE = hf, where EE is energy, hh is Planck's constant (6.626×10−346.626 \times 10^{-34} J·s), and ff is frequency. To find the frequency range, you need to solve for ff using f=Ehf = \frac{E}{h} for both energy limits. For the lower energy limit: fmin=1.0×10−196.626×10−34=1.51×1014f_{min} = \frac{1.0 \times 10^{-19}}{6.626 \times 10^{-34}} = 1.51 \times 10^{14} Hz. For the upper energy limit: fmax=5.0×10−196.626×10−34=7.55×1014f_{max} = \frac{5.0 \times 10^{-19}}{6.626 \times 10^{-34}} = 7.55 \times 10^{14} Hz. This gives us a range of approximately 1.5×10141.5 \times 10^{14} Hz to 7.5×10147.5 \times 10^{14} Hz. Answer A matches this calculation perfectly. Answer B shows frequencies that are too high, suggesting someone might have used an incorrect value for Planck's constant or made a calculation error. Answer C falls in the middle range but doesn't match either boundary correctly. Answer D shows frequencies that are an order of magnitude too low, indicating a possible error in scientific notation or using the wrong constant entirely. Remember that energy and frequency are directly proportional through Planck's constant. Always double-check that you're using the correct value for hh and watch your scientific notation carefully—small errors in exponents lead to dramatically wrong answers in quantum calculations.
  4. The momentum of a photon can be calculated using p=Ecp = \frac{E}{c}. A green laser emits photons with wavelength 532 nm. What is the momentum of each photon?
    Answer: 1.24×10−271.24 \times 10^{-27} kg·m/s This question tests your understanding of photon properties and the relationship between energy, momentum, and wavelength for electromagnetic radiation. When you see photon momentum problems, remember that photons are massless particles, so you can't use the classical momentum formula p=mvp = mv. Instead, you need to connect energy and momentum through the speed of light. To find the photon momentum, start with the given formula p=Ecp = \frac{E}{c}, but you'll need to find the energy first. Use the relationship E=hcλE = \frac{hc}{\lambda}, where h=6.626×10−34h = 6.626 \times 10^{-34} J·s and λ=532\lambda = 532 nm =532×10−9= 532 \times 10^{-9} m. Calculate the energy: E=(6.626×10−34)(3.00×108)532×10−9=3.73×10−19E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{532 \times 10^{-9}} = 3.73 \times 10^{-19} J Now find momentum: p=Ec=3.73×10−193.00×108=1.24×10−27p = \frac{E}{c} = \frac{3.73 \times 10^{-19}}{3.00 \times 10^8} = 1.24 \times 10^{-27} kg·m/s Answer A (1.24×10−271.24 \times 10^{-27} kg·m/s) is correct. Answer B (3.73×10−273.73 \times 10^{-27} kg·m/s) represents the energy value in the wrong units—a common trap when students confuse energy and momentum calculations. Answer C (6.21×10−286.21 \times 10^{-28} kg·m/s) likely results from calculation errors in the energy step. Answer D (1.87×10−261.87 \times 10^{-26} kg·m/s) is an order of magnitude too large, possibly from unit conversion mistakes. Study tip: For photon problems, always remember the connection: wavelength → energy (via E=hc/λE = hc/\lambda) → momentum (via p=E/cp = E/c). Keep your units consistent throughout.
  5. Which statement correctly describes the relationship between photon energy and electromagnetic radiation properties?
    Answer: Energy is inversely proportional to wavelength and directly proportional to frequency When you encounter questions about electromagnetic radiation and photon energy, you're dealing with fundamental quantum mechanics relationships that govern how light and energy interact. The key relationship is Planck's equation: E=hfE = hf, where energy (E) equals Planck's constant (h) times frequency (f). Since the speed of light c=λfc = \lambda f (wavelength times frequency), we can substitute to get E=hcλE = \frac{hc}{\lambda}. These equations reveal that energy increases as frequency increases (direct proportion) and energy decreases as wavelength increases (inverse proportion). Think about this practically: high-energy gamma rays have very high frequencies and very short wavelengths, while low-energy radio waves have low frequencies and long wavelengths. This confirms that option B correctly describes both relationships. Option A reverses both relationships—it incorrectly suggests that longer wavelengths carry more energy, which would mean radio waves are more energetic than X-rays. Option C claims energy increases with both wavelength and frequency simultaneously, but since wavelength and frequency are themselves inversely related (c=λfc = \lambda f), this is mathematically impossible. Option D suggests energy decreases as both wavelength and frequency increase, which again violates the inverse relationship between wavelength and frequency. Remember this pattern: high frequency = short wavelength = high energy. When studying electromagnetic radiation, always connect these three properties together. This relationship appears frequently in chemistry when discussing atomic spectra, photoelectric effects, and molecular transitions.
  6. Which color of visible light photons carries the most energy per photon?
    Answer: Violet light (wavelength ≈ 400 nm) carries the most energy per photon When you encounter questions about photon energy and light color, remember that energy and wavelength have an inverse relationship described by Planck's equation: E=hcλE = \frac{hc}{\lambda}, where h is Planck's constant, c is the speed of light, and λ is wavelength. Since energy is inversely proportional to wavelength, shorter wavelengths carry more energy per photon. Looking at the visible spectrum, violet/blue light has the shortest wavelength (~400-450 nm), followed by green (~530 nm), yellow (~580 nm), and red (~700 nm). However, there's a critical issue with this question: none of the given options (A through D) correctly identifies the highest-energy visible light. Blue light (option D) does have more energy than green, yellow, or red light, but violet light—with wavelengths around 400 nm—actually carries the most energy per photon in the visible spectrum. Option A incorrectly suggests red light has the most energy, when it actually has the least due to its longest wavelength. Option B (yellow) and option C (green) fall in the middle range but still have longer wavelengths than violet. Option D (blue) is closer to correct but still doesn't represent the highest-energy visible light. The answer appears to be E, likely indicating that none of the provided color choices correctly identifies the most energetic visible light photons. Study tip: Remember the mnemonic "Roy G. Biv" for the visible spectrum, and that energy increases as you move from red to violet—shorter wavelength means higher energy.
  7. An X-ray photon has wavelength 1.5×10−101.5 \times 10^{-10} m. How many times more energetic is this photon compared to visible light with wavelength 600 nm?
    Answer: 4000 times more energetic When comparing photon energies, you need to understand the relationship between energy and wavelength. According to Planck's equation, photon energy is E=hcλE = \frac{hc}{\lambda}, which means energy is inversely proportional to wavelength—shorter wavelengths correspond to higher energies. To find how many times more energetic the X-ray photon is, you can set up a ratio: EX−rayEvisible=λvisibleλX−ray\frac{E_{X-ray}}{E_{visible}} = \frac{\lambda_{visible}}{\lambda_{X-ray}}. Notice how the wavelengths flip in the ratio due to the inverse relationship. Converting the visible light wavelength to meters: 600 nm = 6.0×10−76.0 \times 10^{-7} m. Now calculate the ratio: 6.0×10−71.5×10−10=6.01.5×10−710−10=4×103=4000\frac{6.0 \times 10^{-7}}{1.5 \times 10^{-10}} = \frac{6.0}{1.5} \times \frac{10^{-7}}{10^{-10}} = 4 \times 10^3 = 4000 The X-ray photon is 4000 times more energetic than visible light, confirming answer C. Looking at the wrong answers: A (250 times) is far too small and suggests an error in the calculation or unit conversion. B (400 times) is exactly one order of magnitude off—likely from forgetting to convert nanometers to meters or making an arithmetic error with the powers of 10. D (40,000 times) overshoots by a factor of 10, possibly from an error in decimal placement during the division. Remember: when comparing photon energies, you can skip the constants and just take the inverse ratio of wavelengths. Always double-check your unit conversions, especially with nanometers to meters.
  8. Two photons are emitted from excited atoms: Photon A has frequency 5.45×10145.45 \times 10^{14} Hz, and Photon B has wavelength 450 nm. Which photon carries more energy, and by what factor?
    Answer: Photon B carries 1.23 times more energy than Photon A When you encounter photon energy problems, remember that photon energy depends on both frequency and wavelength through Planck's equation: E=hf=hcλE = hf = \frac{hc}{\lambda}, where h=6.626×10−34h = 6.626 \times 10^{-34} J·s and c=3.00×108c = 3.00 \times 10^8 m/s. To compare these photons, you need to calculate each energy. For Photon A with frequency 5.45×10145.45 \times 10^{14} Hz: EA=hf=(6.626×10−34)(5.45×1014)=3.61×10−19E_A = hf = (6.626 \times 10^{-34})(5.45 \times 10^{14}) = 3.61 \times 10^{-19} J For Photon B with wavelength 450 nm (4.50×10−74.50 \times 10^{-7} m): EB=hcλ=(6.626×10−34)(3.00×108)4.50×10−7=4.42×10−19E_B = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{4.50 \times 10^{-7}} = 4.42 \times 10^{-19} J The ratio is EBEA=4.42×10−193.61×10−19=1.22≈1.23\frac{E_B}{E_A} = \frac{4.42 \times 10^{-19}}{3.61 \times 10^{-19}} = 1.22 \approx 1.23 This confirms that Photon B carries 1.23 times more energy than Photon A, making answer B correct. Answer A incorrectly states that Photon A has more energy—it actually has less. Answers C and D use the wrong factor of 2.46, which might result from calculation errors like squaring a conversion factor or confusing the relationship between energy and wavelength/frequency. Study tip: Always convert wavelength to meters before calculating, and remember that higher frequency (shorter wavelength) means higher energy. Blue/violet light (shorter wavelengths around 450 nm) carries more energy than red light (longer wavelengths).
  9. A laser emits 2.5×10182.5 \times 10^{18} photons per second, each with energy 3.0×10−193.0 \times 10^{-19} J. What is the power output of this laser in watts?
    Answer: 0.75 W This question tests your understanding of power calculations involving photons, a fundamental concept in quantum mechanics and laser physics. Power is the rate of energy transfer, measured in watts (joules per second). To find the laser's power output, you need to calculate the total energy emitted per second. Since each photon carries 3.0×10−193.0 \times 10^{-19} J of energy and the laser emits 2.5×10182.5 \times 10^{18} photons per second, multiply these values: Power = (number of photons per second) × (energy per photon) Power = (2.5×1018 photons/s)×(3.0×10−19 J/photon)(2.5 \times 10^{18} \text{ photons/s}) \times (3.0 \times 10^{-19} \text{ J/photon}) Power = 7.5×10−1=0.757.5 \times 10^{-1} = 0.75 W This confirms answer A is correct. Let's examine why the other options are wrong. Answer B (1.2 W) might result from calculation errors in scientific notation manipulation. Answer C (7.5 W) represents a common mistake where students forget to properly handle the exponents—you might get this if you incorrectly multiply 1018×10−19=10110^{18} \times 10^{-19} = 10^{1} instead of 10−110^{-1}. Answer D (12 W) could arise from multiple computational errors or misunderstanding the relationship between the given quantities. Study tip: When working with photon energy problems, always set up your calculation dimensionally first. Check that photons/second × energy/photon gives you energy/second (watts). This dimensional analysis catches many errors before you even calculate, especially when dealing with scientific notation.

Start this quiz