AP Physics 2 Flashcards: The Photoelectric Effect

What this deck covers

This deck focuses on The Photoelectric Effect, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.

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.

Flashcard 1: What is the kinetic energy of ejected electrons?

Answer: $E_k = hf - \text{work function}$. Excess photon energy after overcoming work function becomes electron motion energy.

Flashcard 2: What is the significance of the threshold frequency $f_0$?

Answer: Minimum frequency needed to eject electrons. Frequency below this value provides insufficient energy for electron ejection.

Flashcard 3: What is the unit of the work function?

Answer: Joules (J). Energy units, since work function represents minimum energy required.

Flashcard 4: What does $E_k$ represent in the photoelectric equation?

Answer: Kinetic energy of emitted electrons. Energy of motion possessed by electrons after photoemission occurs.

Flashcard 5: If $E_k = 0$, what condition holds in the photoelectric equation?

Answer: $hf = \text{work function}$. Threshold condition where photon energy exactly equals work function.

Flashcard 6: What is the unit of frequency?

Answer: Hertz (Hz). Standard SI unit for oscillations per second in wave phenomena.

Flashcard 7: What is the relation between stopping potential and maximum kinetic energy?

Answer: Directly proportional. Higher electron kinetic energy requires proportionally higher stopping voltage.

Flashcard 8: Convert $2 \text{ eV}$ to joules.

Answer: $3.20 \times 10^{-19} \text{ J}$. Using conversion $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

Flashcard 9: Identify one factor that does not affect the kinetic energy of emitted electrons.

Answer: Light intensity. Intensity affects electron quantity, not individual electron energy.

Flashcard 10: What is the result of increasing the stopping potential?

Answer: Reduces current to zero. Reverse voltage prevents electrons from reaching collector electrode.

Flashcard 11: What is the effect of increasing light frequency on emitted electrons?

Answer: Increases kinetic energy of electrons. Higher frequency means more energetic photons and faster emitted electrons.

Flashcard 12: What is the result of increasing the stopping potential?

Answer: Reduces current to zero. Reverse voltage prevents electrons from reaching collector electrode.

Flashcard 13: What happens if light intensity increases but frequency is constant?

Answer: More electrons emitted; kinetic energy unchanged. More photons increase electron quantity while frequency determines individual energy.

Flashcard 14: What is the work function in the photoelectric effect?

Answer: Minimum energy needed to remove an electron from a material. Material-specific binding energy threshold for electron emission.

Flashcard 15: Which equation relates energy and frequency of a photon?

Answer: $E = hf$. Planck's relation defining quantized energy packets of electromagnetic radiation.

Flashcard 16: In the photoelectric effect, what is the role of photon energy?

Answer: Provides energy to eject electrons. Photon transfers its energy to overcome binding forces and accelerate electrons.

Flashcard 17: Calculate stopping potential for $E_k = 3 \text{ eV}$.

Answer: $V_s = 3 \text{ V}$. Maximum kinetic energy in eV numerically equals stopping potential in volts.

Flashcard 18: What is the relation between stopping potential and maximum kinetic energy?

Answer: Directly proportional. Higher electron kinetic energy requires proportionally higher stopping voltage.

Flashcard 19: What is the photoelectric effect?

Answer: Emission of electrons from a material when light shines on it. Light energy ejects electrons when photon energy exceeds material's binding energy.

Flashcard 20: What is the work function in the photoelectric effect?

Answer: Minimum energy needed to remove an electron from a material. Material-specific binding energy threshold for electron emission.

Flashcard 21: What does $E_k$ represent in the photoelectric equation?

Answer: Kinetic energy of emitted electrons. Energy of motion possessed by electrons after photoemission occurs.

Flashcard 22: What happens if the light frequency is below the threshold frequency?

Answer: No electrons are emitted. Insufficient photon energy cannot overcome the material's work function.

Flashcard 23: Which constant is $6.626 \times 10^{-34} \text{ J s}$?

Answer: Planck's constant. Fundamental quantum constant relating energy to frequency in photon interactions.

Flashcard 24: What is the photoelectric work function symbol?

Answer: $\text{work function}$ or $\text{W}$. Standard symbols representing minimum energy for electron removal from material.

Flashcard 25: What is the photoelectric work function symbol?

Answer: $\text{work function}$ or $\text{W}$. Standard symbols representing minimum energy for electron removal from material.

Flashcard 26: What happens if the light frequency is below the threshold frequency?

Answer: No electrons are emitted. Insufficient photon energy cannot overcome the material's work function.

Flashcard 27: Calculate frequency for wavelength $500 \text{ nm}$.

Answer: $f = 6.00 \times 10^{14} \text{ Hz}$. Using $f = c/\lambda$ with $\lambda = 500 \times 10^{-9} \text{ m}$.

Flashcard 28: What is the effect of a higher work function on electron emission?

Answer: Requires higher frequency light to eject electrons. Greater binding energy demands more energetic photons for electron liberation.

Flashcard 29: What is the unit of Planck's constant?

Answer: Joule-seconds (J s). Energy-time units reflecting quantum action in electromagnetic interactions.

Flashcard 30: What is the effect of increasing light intensity on electron emission?

Answer: Increases number of electrons emitted. More photons hit surface, but individual electron energy remains constant.

Flashcard 31: State Einstein's photoelectric equation.

Answer: $E_k = hf - \text{work function}$. Photon energy minus work function equals electron's kinetic energy.

Flashcard 32: How does light frequency affect the kinetic energy of electrons?

Answer: Higher frequency increases kinetic energy. More energetic photons transfer greater kinetic energy to ejected electrons.

Flashcard 33: What is the unit of frequency?

Answer: Hertz (Hz). Standard SI unit for oscillations per second in wave phenomena.

Flashcard 34: Identify the threshold frequency symbol.

Answer: $f_0$. Standard notation for minimum frequency required for electron emission.

Flashcard 35: Calculate the photon energy for $f = 6 \times 10^{14} \text{ Hz}$.

Answer: $E = 3.98 \times 10^{-19} \text{ J}$. Using $E = hf$ with $h = 6.626 \times 10^{-34} \text{ J s}$.

Flashcard 36: What is the relationship between frequency and energy of light?

Answer: Directly proportional. Higher frequency photons carry more energy: $E = hf$.

Flashcard 37: What happens if light intensity increases but frequency is constant?

Answer: More electrons emitted; kinetic energy unchanged. More photons increase electron quantity while frequency determines individual energy.

Flashcard 38: Which constant is $6.626 \times 10^{-34} \text{ J s}$?

Answer: Planck's constant. Fundamental quantum constant relating energy to frequency in photon interactions.

Flashcard 39: State the relationship between photon frequency and wavelength.

Answer: $\text{Frequency} \times \text{Wavelength} = c$. Wave equation $c = f\lambda$ relates electromagnetic wave properties.

Flashcard 40: What is the effect of increasing light intensity on electron emission?

Answer: Increases number of electrons emitted. More photons hit surface, but individual electron energy remains constant.

Flashcard 41: How does light frequency affect the kinetic energy of electrons?

Answer: Higher frequency increases kinetic energy. More energetic photons transfer greater kinetic energy to ejected electrons.

Flashcard 42: What is the speed of light in vacuum?

Answer: $c = 3.00 \times 10^8 \text{ m/s}$. Fundamental constant for electromagnetic wave propagation in vacuum.

Flashcard 43: Calculate stopping potential for $E_k = 3 \text{ eV}$.

Answer: $V_s = 3 \text{ V}$. Maximum kinetic energy in eV numerically equals stopping potential in volts.

Flashcard 44: What effect does increasing light intensity have on photocurrent?

Answer: Increases photocurrent. More photons create greater electron flow in photoelectric circuit.

Flashcard 45: What is the kinetic energy of ejected electrons?

Answer: $E_k = hf - \text{work function}$. Excess photon energy after overcoming work function becomes electron motion energy.

Flashcard 46: Identify the effect of increasing voltage on the photoelectric current.

Answer: Increases current until saturation. Forward voltage accelerates electrons until all available electrons flow.

Flashcard 47: What does $h$ represent in the photoelectric equation?

Answer: Planck's constant. Fundamental constant linking energy and frequency: $h = 6.626 \times 10^{-34} \text{ J s}$.

Flashcard 48: What is the significance of the threshold frequency $f_0$?

Answer: Minimum frequency needed to eject electrons. Frequency below this value provides insufficient energy for electron ejection.

Flashcard 49: In the photoelectric effect, what is the role of photon energy?

Answer: Provides energy to eject electrons. Photon transfers its energy to overcome binding forces and accelerate electrons.

Flashcard 50: What is the effect of a higher work function on electron emission?

Answer: Requires higher frequency light to eject electrons. Greater binding energy demands more energetic photons for electron liberation.

Flashcard 51: Find the maximum kinetic energy if $hf = 5 \text{ eV}$ and $\text{work function} = 2 \text{ eV}$.

Answer: $E_k = 3 \text{ eV}$. Applying Einstein's equation: $E_k = 5 - 2 = 3 \text{ eV}$.

Flashcard 52: Convert $2 \text{ eV}$ to joules.

Answer: $3.20 \times 10^{-19} \text{ J}$. Using conversion $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

Flashcard 53: What effect does increasing light intensity have on photocurrent?

Answer: Increases photocurrent. More photons create greater electron flow in photoelectric circuit.

Flashcard 54: What is the unit of the work function?

Answer: Joules (J). Energy units, since work function represents minimum energy required.

Flashcard 55: Identify one factor that does not affect the kinetic energy of emitted electrons.

Answer: Light intensity. Intensity affects electron quantity, not individual electron energy.

Flashcard 56: What is the relationship between frequency and energy of light?

Answer: Directly proportional. Higher frequency photons carry more energy: $E = hf$.

Flashcard 57: Find the work function if $hf = 4 \text{ eV}$ and $E_k = 1 \text{ eV}$.

Answer: $\text{work function} = 3 \text{ eV}$. Rearranging Einstein's equation: work function $= hf - E_k$.

Flashcard 58: Calculate the photon energy for $f = 6 \times 10^{14} \text{ Hz}$.

Answer: $E = 3.98 \times 10^{-19} \text{ J}$. Using $E = hf$ with $h = 6.626 \times 10^{-34} \text{ J s}$.

Flashcard 59: Which law does the photoelectric effect contradict in classical physics?

Answer: Wave theory of light. Classical theory predicted continuous energy dependence on intensity, not frequency.

Flashcard 60: Which law does the photoelectric effect contradict in classical physics?

Answer: Wave theory of light. Classical theory predicted continuous energy dependence on intensity, not frequency.

Flashcard 61: What does $h$ represent in the photoelectric equation?

Answer: Planck's constant. Fundamental constant linking energy and frequency: $h = 6.626 \times 10^{-34} \text{ J s}$.

Flashcard 62: What is the effect of increasing light frequency on emitted electrons?

Answer: Increases kinetic energy of electrons. Higher frequency means more energetic photons and faster emitted electrons.

Flashcard 63: Find the work function if $hf = 4 \text{ eV}$ and $E_k = 1 \text{ eV}$.

Answer: $\text{work function} = 3 \text{ eV}$. Rearranging Einstein's equation: work function $= hf - E_k$.

Flashcard 64: Which equation relates energy and frequency of a photon?

Answer: $E = hf$. Planck's relation defining quantized energy packets of electromagnetic radiation.

Flashcard 65: State Einstein's photoelectric equation.

Answer: $E_k = hf - \text{work function}$. Photon energy minus work function equals electron's kinetic energy.

Flashcard 66: Calculate frequency for wavelength $500 \text{ nm}$.

Answer: $f = 6.00 \times 10^{14} \text{ Hz}$. Using $f = c/\lambda$ with $\lambda = 500 \times 10^{-9} \text{ m}$.

Flashcard 67: If $E_k = 0$, what condition holds in the photoelectric equation?

Answer: $hf = \text{work function}$. Threshold condition where photon energy exactly equals work function.

Flashcard 68: What is the speed of light in vacuum?

Answer: $c = 3.00 \times 10^8 \text{ m/s}$. Fundamental constant for electromagnetic wave propagation in vacuum.

Flashcard 69: State the equation for stopping potential $V_s$.

Answer: $eV_s = E_k$. Voltage required to stop fastest electrons equals their kinetic energy.

Flashcard 70: Find the maximum kinetic energy if $hf = 5 \text{ eV}$ and $\text{work function} = 2 \text{ eV}$.

Answer: $E_k = 3 \text{ eV}$. Applying Einstein's equation: $E_k = 5 - 2 = 3 \text{ eV}$.

Flashcard 71: State the relationship between photon frequency and wavelength.

Answer: $\text{Frequency} \times \text{Wavelength} = c$. Wave equation $c = f\lambda$ relates electromagnetic wave properties.

Flashcard 72: Identify the effect of increasing voltage on the photoelectric current.

Answer: Increases current until saturation. Forward voltage accelerates electrons until all available electrons flow.

Flashcard 73: Identify the threshold frequency symbol.

Answer: $f_0$. Standard notation for minimum frequency required for electron emission.