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The photoelectric effect is the phenomenon where electrons are emitted from a material when light is shone on it. This effect is crucial for understanding the particle nature of light and the concept of threshold frequency.
This topic appears in exams because it tests your understanding of quantum mechanics and the interaction between light and matter. Questions typically involve calculating the kinetic energy of emitted electrons or determining the threshold frequency.
This topic is frequently tested in physics exams, especially in high school and undergraduate levels. It carries significant marks and tests your ability to apply fundamental quantum principles. Understanding the photoelectric effect is essential for fields like optics, semiconductor technology, and solar energy.
The kinetic energy (KE) of the ejected electron is given by Einstein's Photoelectric Equation: [ KE = hf - \phi ] where: - h is Planck's constant - f is the frequency of the incident light - ? is the work function of the material
Imagine a staircase where each step represents the energy required to eject an electron. The height of the staircase is the work function (?), and the energy of the incident photon (hf) must be at least as high as the top step to eject an electron.
Intermediate
Question: Calculate the kinetic energy of an electron ejected from a material with a work function of 2.0 eV when light of frequency 6.0 × 10^14 Hz is incident on it. (Planck's constant h = 6.626 × 10^-34 J·s)
Step-by-Step:1. Convert the work function to Joules: ( \phi = 2.0 \, eV \times 1.6 \times 10^{-19} \, J/eV = 3.2 \times 10^{-19} \, J )2. Calculate the energy of the incident photon: ( E = hf = 6.626 \times 10^{-34} \, J·s \times 6.0 \times 10^{14} \, Hz = 3.98 \times 10^{-19} \, J )3. Use Einstein's equation: ( KE = E - \phi = 3.98 \times 10^{-19} \, J - 3.2 \times 10^{-19} \, J = 0.78 \times 10^{-19} \, J )
Answer: ( KE = 0.78 \times 10^{-19} \, J )
Question: Determine the threshold frequency for a material with a work function of 2.5 eV.
Step-by-Step:1. Convert the work function to Joules: ( \phi = 2.5 \, eV \times 1.6 \times 10^{-19} \, J/eV = 4.0 \times 10^{-19} \, J )2. Use the threshold frequency formula: ( f_0 = \frac{\phi}{h} = \frac{4.0 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, J·s} = 6.04 \times 10^{14} \, Hz )
Answer: ( f_0 = 6.04 \times 10^{14} \, Hz )
Question: Light of wavelength 400 nm is incident on a material with a work function of 2.2 eV. Calculate the kinetic energy of the ejected electrons.
Step-by-Step:1. Convert the work function to Joules: ( \phi = 2.2 \, eV \times 1.6 \times 10^{-19} \, J/eV = 3.52 \times 10^{-19} \, J )2. Calculate the frequency of the incident light: ( f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, m/s}{400 \times 10^{-9} \, m} = 7.5 \times 10^{14} \, Hz )3. Calculate the energy of the incident photon: ( E = hf = 6.626 \times 10^{-34} \, J·s \times 7.5 \times 10^{14} \, Hz = 4.97 \times 10^{-19} \, J )4. Use Einstein's equation: ( KE = E - \phi = 4.97 \times 10^{-19} \, J - 3.52 \times 10^{-19} \, J = 1.45 \times 10^{-19} \, J )
Answer: ( KE = 1.45 \times 10^{-19} \, J )
Correct Approach: Always convert eV to Joules using ( 1 \, eV = 1.6 \times 10^{-19} \, J ).
Mistake: Confusing frequency and wavelength.
Correct Approach: Convert wavelength to frequency using ( f = \frac{c}{\lambda} ).
Mistake: Not understanding the threshold frequency concept.
Correct Approach: Electrons are only ejected if the frequency is greater than the threshold frequency.
Mistake: Incorrectly applying Einstein's equation.
Favored By: Physics exams, engineering tests.
Conceptual Questions: Explain the photoelectric effect or define threshold frequency.
Favored By: High school physics, conceptual tests.
Multiple-Choice Questions: Identify the correct formula or concept.
Question: What is the kinetic energy of an electron ejected from a material with a work function of 2.0 eV when light of frequency 6.0 × 10^14 Hz is incident on it? (Planck's constant h = 6.626 × 10^-34 J·s) - A: 0.78 × 10^-19 J - B: 3.2 × 10^-19 J - C: 3.98 × 10^-19 J - D: 4.0 × 10^-19 J
Correct Answer: A Explanation: Convert the work function to Joules and use Einstein's equation. Why the Distractors Are Tempting: B is the work function in Joules, C is the photon energy, D is a random value.
Question: Determine the threshold frequency for a material with a work function of 2.5 eV. - A: 6.04 × 10^14 Hz - B: 4.0 × 10^14 Hz - C: 2.5 × 10^14 Hz - D: 3.98 × 10^14 Hz
Correct Answer: A Explanation: Convert the work function to Joules and use the threshold frequency formula. Why the Distractors Are Tempting: B and C are random values, D is a common photon energy.
Question: Light of wavelength 400 nm is incident on a material with a work function of 2.2 eV. Calculate the kinetic energy of the ejected electrons. - A: 1.45 × 10^-19 J - B: 3.52 × 10^-19 J - C: 4.97 × 10^-19 J - D: 2.2 × 10^-19 J
Correct Answer: A Explanation: Convert the work function to Joules, calculate the frequency, and use Einstein's equation. Why the Distractors Are Tempting: B is the work function in Joules, C is the photon energy, D is a random value.
Question: Which of the following is the correct formula for the kinetic energy of an ejected electron? - A: KE = hf + ? - B: KE = hf - ? - C: KE = hf / ? - D: KE = hf * ?
Correct Answer: B Explanation: Einstein's photoelectric equation. Why the Distractors Are Tempting: A is incorrect addition, C and D are incorrect operations.
Question: If the frequency of the incident light is below the threshold frequency, what happens? - A: Electrons are ejected with less kinetic energy. - B: Electrons are ejected with more kinetic energy. - C: No electrons are ejected. - D: The work function increases.
Correct Answer: C Explanation: Below the threshold frequency, no electrons are ejected. Why the Distractors Are Tempting: A and B suggest incorrect kinetic energy changes, D is an irrelevant change.
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