AP Chemistry Flashcards: Properties Of Photons

Study Properties Of Photons in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Chemistry

Properties Of Photons

0 mastered0 still learning

0% Complete

QUESTION
1/ 64

Which fundamental constant is used in calculating photon's energy?

Tap card or press Space to flip

ANSWER

Planck's constant (hh). Quantum constant linking energy and frequency in photon equations.

How well did you know it?

Card 1 / 64

What this deck covers

This deck focuses on Properties Of Photons, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

All flashcards

Flashcard 1: Which fundamental constant is used in calculating photon's energy?

Answer: Planck's constant (hh). Quantum constant linking energy and frequency in photon equations.

Flashcard 2: What is the relationship between photon momentum and wavelength?

Answer: p=hλp = \frac{h}{\lambda}. Momentum inversely related to wavelength via Planck's constant.

Flashcard 3: Calculate energy of a photon with wavelength 400 nm400 \text{ nm}.

Answer: E=4.97×1019 JE = 4.97 \times 10^{-19} \text{ J}. Using E=hcλE = \frac{hc}{\lambda} with 400400 nm wavelength.

Flashcard 4: Convert energy of 1.5×10191.5 \times 10^{-19} J to electronvolts.

Answer: 0.9375 eV0.9375 \text{ eV}. Converting joules to eV using standard conversion factor.

Flashcard 5: Find the energy of a photon with frequency 5×10145 \times 10^{14} Hz.

Answer: E=3.31×1019 JE = 3.31 \times 10^{-19} \text{ J}. Using E=hfE = hf with h=6.626×1034h = 6.626 \times 10^{-34} J·s.

Flashcard 6: What is the relationship between photon momentum and wavelength?

Answer: p=hλp = \frac{h}{\lambda}. Momentum inversely related to wavelength via Planck's constant.

Flashcard 7: Identify the formula for the momentum of a photon.

Answer: p=hwavelengthp = \frac{h}{\text{wavelength}}. De Broglie relation for photon momentum using wavelength.

Flashcard 8: State the unit for frequency of a photon.

Answer: Hertz (Hz). SI unit measuring cycles per second.

Flashcard 9: Determine the frequency of a photon with a wavelength of 500 nm500 \text{ nm}.

Answer: ν=6.00×1014 Hz\nu = 6.00 \times 10^{14} \text{ Hz}. Using ν=cλ\nu = \frac{c}{\lambda} with 500500 nm wavelength.

Flashcard 10: How do you express energy in electronvolts (eV)?

Answer: 1 eV = 1.602×10191.602 \times 10^{-19} J. Conversion factor between joules and electronvolts.

Flashcard 11: Determine energy in eV for photon with 4×10194 \times 10^{-19} J.

Answer: 2.5 eV2.5 \text{ eV}. Converting joules to eV using conversion factor.

Flashcard 12: If a photon's wavelength is halved, what happens to its energy?

Answer: Energy doubles. Energy inversely proportional to wavelength per E=hcλE = \frac{hc}{\lambda}.

Flashcard 13: Calculate frequency for photon with energy 2.5×10192.5 \times 10^{-19} J.

Answer: ν=3.77×1014 Hz\nu = 3.77 \times 10^{14} \text{ Hz}. Using ν=Eh\nu = \frac{E}{h} with given energy value.

Flashcard 14: What does the symbol λ\lambda represent?

Answer: Wavelength. Lambda is the standard symbol for wavelength.

Flashcard 15: What is the relationship between photon momentum and wavelength?

Answer: p=hλp = \frac{h}{\lambda}. Momentum inversely related to wavelength via Planck's constant.

Flashcard 16: Calculate frequency for photon with energy 2.5×10192.5 \times 10^{-19} J.

Answer: ν=3.77×1014 Hz\nu = 3.77 \times 10^{14} \text{ Hz}. Using ν=Eh\nu = \frac{E}{h} with given energy value.

Flashcard 17: Calculate the frequency for a photon with 3×10193 \times 10^{-19} J energy.

Answer: ν=4.53×1014 Hz\nu = 4.53 \times 10^{14} \text{ Hz}. Using ν=Eh\nu = \frac{E}{h} with given energy.

Flashcard 18: What is the primary property that determines a photon's energy?

Answer: Frequency. Higher frequency means higher energy per Planck's equation.

Flashcard 19: Calculate energy of a photon with wavelength 400 nm400 \text{ nm}.

Answer: E=4.97×1019 JE = 4.97 \times 10^{-19} \text{ J}. Using E=hcλE = \frac{hc}{\lambda} with 400400 nm wavelength.

Flashcard 20: State the unit for photon energy.

Answer: Joules (J). Standard SI unit for energy in all physics contexts.

Flashcard 21: What is Planck's equation for energy?

Answer: E=hfE = hf. Fundamental quantum equation for photon energy.

Flashcard 22: What is the relationship between wavelength and energy?

Answer: Inversely proportional. Energy increases as wavelength decreases per E=hcλE = \frac{hc}{\lambda}.

Flashcard 23: What is the speed of light in a vacuum?

Answer: c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}. Universal constant for electromagnetic radiation in vacuum.

Flashcard 24: Find the energy of a photon with frequency 5×10145 \times 10^{14} Hz.

Answer: E=3.31×1019 JE = 3.31 \times 10^{-19} \text{ J}. Using E=hfE = hf with h=6.626×1034h = 6.626 \times 10^{-34} J·s.

Flashcard 25: State the symbol for speed of light.

Answer: cc. Standard symbol for speed of light constant.

Flashcard 26: Calculate energy of a photon with wavelength 400 nm400 \text{ nm}.

Answer: E=4.97×1019 JE = 4.97 \times 10^{-19} \text{ J}. Using E=hcλE = \frac{hc}{\lambda} with 400400 nm wavelength.

Flashcard 27: What is the inverse relationship of wavelength and frequency?

Answer: As wavelength increases, frequency decreases. Inverse relationship from c=νλc = \nu \lambda equation.

Flashcard 28: What unit is used for Planck's constant?

Answer: Joule seconds (J s). Action unit combining energy and time dimensions.

Flashcard 29: What is the equation for the energy of a photon?

Answer: E=hfE = hf. Planck's equation relating energy to frequency with constant hh.

Flashcard 30: What is the inverse relationship of wavelength and frequency?

Answer: As wavelength increases, frequency decreases. Inverse relationship from c=νλc = \nu \lambda equation.

Flashcard 31: Find the energy of a photon with frequency 5×10145 \times 10^{14} Hz.

Answer: E=3.31×1019 JE = 3.31 \times 10^{-19} \text{ J}. Using E=hfE = hf with h=6.626×1034h = 6.626 \times 10^{-34} J·s.

Flashcard 32: Identify the unit for measuring photon wavelength.

Answer: Meters (m). Standard SI unit for measuring distance and wavelength.

Flashcard 33: Determine the energy in eV for a photon with 3.2×10193.2 \times 10^{-19} J.

Answer: 2 eV. Dividing energy by conversion factor 1.602×10191.602 \times 10^{-19} J/eV.

Flashcard 34: What is the relationship between frequency and photon energy?

Answer: Directly proportional. Linear relationship from Planck's equation E=hfE = hf.

Flashcard 35: What is the formula for photon frequency in terms of speed and wavelength?

Answer: ν=cλ\nu = \frac{c}{\lambda}. Frequency equals speed of light divided by wavelength.

Flashcard 36: State the formula relating frequency and wavelength of a photon.

Answer: c=ν×wavelengthc = \nu \times \text{wavelength}. Speed of light equals frequency times wavelength.

Flashcard 37: Determine the energy in eV for a photon with 3.2×10193.2 \times 10^{-19} J.

Answer: 2 eV. Dividing energy by conversion factor 1.602×10191.602 \times 10^{-19} J/eV.

Flashcard 38: What does the symbol ν\nu represent?

Answer: Frequency. Nu is the standard symbol for frequency.

Flashcard 39: What is the symbol for frequency?

Answer: ν\nu (nu). Greek letter nu represents frequency in physics equations.

Flashcard 40: Which fundamental constant is used in calculating photon's energy?

Answer: Planck's constant (hh). Quantum constant linking energy and frequency in photon equations.

Flashcard 41: Find momentum of a photon with wavelength 700 nm700 \text{ nm}.

Answer: p=9.47×1028 kg m/sp = 9.47 \times 10^{-28} \text{ kg m/s}. Using p=hλp = \frac{h}{\lambda} with 700700 nm wavelength.

Flashcard 42: Which constant represents Planck's constant?

Answer: h=6.626×1034 J sh = 6.626 \times 10^{-34} \text{ J s}. Fundamental constant relating energy to frequency in quantum mechanics.

Flashcard 43: What is the relationship between energy and frequency?

Answer: E=hfE = hf. Energy increases linearly with frequency via Planck's constant.

Flashcard 44: What does the energy of a photon depend on?

Answer: Frequency. Energy is directly proportional to frequency only.

Flashcard 45: What does the symbol λ\lambda represent?

Answer: Wavelength. Lambda is the standard symbol for wavelength.

Flashcard 46: What is the inverse relationship of wavelength and frequency?

Answer: As wavelength increases, frequency decreases. Inverse relationship from c=νλc = \nu \lambda equation.

Flashcard 47: What does the symbol λ\lambda represent?

Answer: Wavelength. Lambda is the standard symbol for wavelength.

Flashcard 48: Choose the symbol for wavelength.

Answer: λ\lambda (lambda). Greek letter lambda represents wavelength in physics.

Flashcard 49: What is the wavelength of a photon with frequency 6×10146 \times 10^{14} Hz?

Answer: λ=5×107 m\lambda = 5 \times 10^{-7} \text{ m}. Using λ=cν\lambda = \frac{c}{\nu} with given frequency.

Flashcard 50: Identify the formula for the momentum of a photon.

Answer: p=hwavelengthp = \frac{h}{\text{wavelength}}. De Broglie relation for photon momentum using wavelength.

Flashcard 51: What is the relationship between frequency and photon energy?

Answer: Directly proportional. Linear relationship from Planck's equation E=hfE = hf.

Flashcard 52: State the symbol for speed of light.

Answer: cc. Standard symbol for speed of light constant.

Flashcard 53: How do you express energy in electronvolts (eV)?

Answer: 1 eV = 1.602×10191.602 \times 10^{-19} J. Conversion factor between joules and electronvolts.

Flashcard 54: What is the relationship between frequency and photon energy?

Answer: Directly proportional. Linear relationship from Planck's equation E=hfE = hf.

Flashcard 55: Which fundamental constant is used in calculating photon's energy?

Answer: Planck's constant (hh). Quantum constant linking energy and frequency in photon equations.

Flashcard 56: How do you express energy in electronvolts (eV)?

Answer: 1 eV = 1.602×10191.602 \times 10^{-19} J. Conversion factor between joules and electronvolts.

Flashcard 57: Calculate the wavelength of a photon with energy 4×10194 \times 10^{-19} J.

Answer: λ=4.97×107 m\lambda = 4.97 \times 10^{-7} \text{ m}. Using λ=hcE\lambda = \frac{hc}{E} with given energy value.

Flashcard 58: Identify the constant used for the speed of light.

Answer: c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}. Symbol cc represents speed of light in equations.

Flashcard 59: Identify the constant used for the speed of light.

Answer: c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}. Symbol cc represents speed of light in equations.

Flashcard 60: Identify the constant used for the speed of light.

Answer: c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s}. Symbol cc represents speed of light in equations.

Flashcard 61: Determine the energy in eV for a photon with 3.2×10193.2 \times 10^{-19} J.

Answer: 2 eV. Dividing energy by conversion factor 1.602×10191.602 \times 10^{-19} J/eV.

Flashcard 62: Calculate frequency for photon with energy 2.5×10192.5 \times 10^{-19} J.

Answer: ν=3.77×1014 Hz\nu = 3.77 \times 10^{14} \text{ Hz}. Using ν=Eh\nu = \frac{E}{h} with given energy value.

Flashcard 63: How is wavelength related to photon energy?

Answer: E=hcwavelengthE = \frac{hc}{\text{wavelength}}. Combines Planck's equation with speed of light relation.

Flashcard 64: Identify the formula for the momentum of a photon.

Answer: p=hwavelengthp = \frac{h}{\text{wavelength}}. De Broglie relation for photon momentum using wavelength.