photoelectric effect problems with answers
Gordon Pfannerstill
photoelectric effect problems with answers are essential for students and enthusiasts aiming to understand the fundamental principles of quantum physics and the behavior of electrons under electromagnetic radiation. The photoelectric effect, first explained by Albert Einstein in 1905, fundamentally changed our understanding of light and matter interaction. This phenomenon involves the emission of electrons from a material—usually a metal surface—when it is exposed to incident light of sufficient frequency. To deepen your grasp of this phenomenon, working through practical problems with solutions can be highly beneficial. This article provides a comprehensive collection of photoelectric effect problems with detailed answers, explanations, and tips to enhance your understanding.
Understanding the Photoelectric Effect
Before diving into problems, it is crucial to understand the basic concepts and formulas related to the photoelectric effect.
Key Concepts
- Work Function (\(\phi\)): The minimum energy required to eject an electron from the surface of a metal.
- Photon Energy (\(E\)): Given by \(E = hf = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(f\) is the frequency, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
- Kinetic Energy of Ejected Electron (\(K.E.\)): The excess energy after overcoming the work function, given by \(K.E. = hf - \phi\).
- Stopping Potential (\(V_s\)): The voltage needed to stop the most energetic ejected electrons, related to their maximum kinetic energy by \(K.E._{\max} = eV_s\), where \(e\) is the elementary charge.
Important Formulas
- Photon energy: \(E = hf = \frac{hc}{\lambda}\)
- Maximum kinetic energy: \(K.E._{\max} = hf - \phi\)
- Stopping potential: \(V_s = \frac{K.E._{\max}}{e} = \frac{hf - \phi}{e}\)
Sample Photoelectric Effect Problems with Answers
Below are some common problems, ranging from basic to advanced, designed to test and reinforce your understanding.
Problem 1: Determining the Work Function
Question: Light of wavelength 500 nm falls on a metal surface. The maximum kinetic energy of emitted electrons is 2 eV. Find the work function of the metal.
Solution:
- Calculate the photon energy:
\[
E = \frac{hc}{\lambda}
\]
Using \(h = 6.626 \times 10^{-34} \text{ Js}\), \(c = 3.0 \times 10^8 \text{ m/s}\), and \(\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}\):
\[
E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{500 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{5 \times 10^{-7}} = 3.9756 \times 10^{-19} \text{ J}
\]
- Convert photon energy to electron volts:
\[
1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}
\]
\[
E = \frac{3.9756 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.48 \text{ eV}
\]
- Use the photoelectric equation:
\[
K.E._{\max} = E - \phi
\]
\[
\phi = E - K.E._{\max} = 2.48 \text{ eV} - 2 \text{ eV} = 0.48 \text{ eV}
\]
Answer:
The work function of the metal is approximately 0.48 eV.
Problem 2: Calculating the Stopping Potential
Question: Monochromatic light of wavelength 600 nm shines on a metal surface. The work function of the metal is 2 eV. Find the stopping potential required to stop the emitted electrons.
Solution:
- Calculate photon energy:
\[
E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{600 \times 10^{-9}} = 3.31 \times 10^{-19} \text{ J}
\]
Convert to eV:
\[
E = \frac{3.31 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.07 \text{ eV}
\]
- Calculate maximum kinetic energy:
\[
K.E._{\max} = E - \phi = 2.07 \text{ eV} - 2 \text{ eV} = 0.07 \text{ eV}
\]
- Find the stopping potential:
\[
V_s = \frac{K.E._{\max}}{e} = 0.07 \text{ V}
\]
Answer:
The stopping potential required is approximately 0.07 V.
Problem 3: Effect of Wavelength on Photoelectric Emission
Question: Explain why increasing the wavelength of incident light reduces the kinetic energy of emitted electrons, assuming the intensity remains constant.
Answer:
Increasing the wavelength (\(\lambda\)) of incident light decreases its photon energy (\(E = \frac{hc}{\lambda}\)) because photon energy is inversely proportional to wavelength. Since the maximum kinetic energy of emitted electrons is given by \(K.E._{\max} = hf - \phi\), a decrease in \(E\) results in a lower \(K.E._{\max}\). If the photon energy becomes less than the work function \(\phi\), no electrons will be emitted regardless of the light's intensity. Therefore, longer wavelengths lead to lower kinetic energies of emitted electrons and can eventually prevent electron emission altogether if the photon energy drops below the work function threshold.
Problem 4: Determining the Frequency Threshold
Question: Find the threshold frequency for a metal with a work function of 1.5 eV.
Solution:
Use the relation:
\[
\phi = hf_{threshold}
\]
Convert work function to joules:
\[
\phi = 1.5 \text{ eV} = 1.5 \times 1.602 \times 10^{-19} = 2.403 \times 10^{-19} \text{ J}
\]
Calculate threshold frequency:
\[
f_{threshold} = \frac{\phi}{h} = \frac{2.403 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 3.63 \times 10^{14} \text{ Hz}
\]
Answer:
The threshold frequency is approximately 3.63 \times 10^{14} Hz.
Problem 5: Impact of Light Intensity on Photoelectric Emission
Question: Does increasing the intensity of incident light increase the maximum kinetic energy of emitted electrons? Justify your answer.
Answer:
No, increasing the intensity of incident light does not increase the maximum kinetic energy of emitted electrons. The maximum kinetic energy depends solely on the photon energy and the work function, given by:
\[
K.E._{\max} = hf - \phi
\]
Since the photon energy \(hf\) is determined by the wavelength (or frequency) of the light, changing the intensity (which relates to the number of photons) does not affect the energy per photon. Therefore, while higher intensity results in more electrons being emitted, the maximum kinetic energy remains unchanged.
Additional Tips for Solving Photoelectric Effect Problems
- Always convert units carefully, especially when dealing with wavelengths, energies, and voltages.
- Remember the inverse relationship between wavelength and photon energy.
- Use the correct constants: Planck's constant (\(6.626 \times 10^{-34}\) Js), speed of
Photoelectric Effect Problems with Answers: An In-Depth Review
The photoelectric effect stands as one of the foundational phenomena that cemented the shift from classical to quantum physics. It not only challenged existing theories but also laid the groundwork for the development of quantum mechanics. As such, understanding the intricacies and potential problem-solving approaches related to the photoelectric effect is crucial for students, educators, and researchers alike. This review aims to present a comprehensive analysis of common photoelectric effect problems, complete with detailed solutions and explanations, fostering a deeper conceptual understanding.
Introduction to the Photoelectric Effect
The photoelectric effect occurs when photons incident on a material surface eject electrons from that surface. The fundamental principles involve the quantization of light energy and the interaction of photons with electrons in a material. The phenomenon is characterized by several key parameters:
- Threshold Frequency (\(f_0\)): The minimum frequency of incident light required to eject electrons.
- Work Function (\(\phi\)): The minimum energy needed to eject an electron from the material, related to the threshold frequency by \(\phi = h f_0\).
- Kinetic Energy of Ejected Electrons: Given by Einstein’s photoelectric equation \(K_{max} = hf - \phi\), where \(hf\) is the photon energy.
Understanding these parameters forms the basis for solving related problems.
Common Photoelectric Effect Problems and Solutions
Below, we explore typical problem types encountered in the context of the photoelectric effect, complete with step-by-step solutions.
Problem Type 1: Calculating the Kinetic Energy of Ejected Electrons
Problem Statement:
A mercury vapor lamp emits light at a wavelength of 253.7 nm. The work function of the photoemissive material is \( \phi = 4.9\, \text{eV} \). What is the maximum kinetic energy of the emitted electrons?
Solution:
Step 1: Convert wavelength to photon energy
Photon energy is given by:
\[
hf = \frac{hc}{\lambda}
\]
Where:
- \(h = 6.626 \times 10^{-34}\, \text{Js}\),
- \(c = 3.0 \times 10^{8}\, \text{m/s}\),
- \(\lambda = 253.7\, \text{nm} = 253.7 \times 10^{-9}\, \text{m}\).
Calculating:
\[
hf = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{253.7 \times 10^{-9}} \approx 7.83 \times 10^{-19}\, \text{J}
\]
Convert Joules to eV (since \(1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\)):
\[
hf \approx \frac{7.83 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.89\, \text{eV}
\]
Step 2: Apply the photoelectric equation
\[
K_{max} = hf - \phi
\]
\[
K_{max} = 4.89\, \text{eV} - 4.9\, \text{eV} = -0.01\, \text{eV}
\]
Since kinetic energy cannot be negative, it indicates that the photon energy is just below the work function, and no electrons are emitted at this wavelength.
Answer:
No electrons are emitted because the photon energy is insufficient to overcome the work function.
Problem Type 2: Determining the Threshold Frequency or Wavelength
Problem Statement:
A metal surface has a work function of 2.3 eV. Find the threshold wavelength for photoemission.
Solution:
Step 1: Convert work function to Joules
\[
\phi = 2.3\, \text{eV} = 2.3 \times 1.602 \times 10^{-19} \approx 3.685 \times 10^{-19}\, \text{J}
\]
Step 2: Calculate threshold frequency
\[
f_0 = \frac{\phi}{h} = \frac{3.685 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 5.56 \times 10^{14}\, \text{Hz}
\]
Step 3: Find threshold wavelength
\[
\lambda_0 = \frac{c}{f_0} = \frac{3.0 \times 10^{8}}{5.56 \times 10^{14}} \approx 539\, \text{nm}
\]
Answer:
Threshold wavelength: approximately 539 nm.
Any incident light with wavelength shorter than 539 nm (higher frequency) can cause photoemission; longer wavelengths cannot.
Problem Type 3: Calculating the Maximum Kinetic Energy for Given Wavelength
Problem Statement:
Light with wavelength 200 nm strikes a metal surface with work function 2.2 eV. What is the maximum kinetic energy of the emitted electrons?
Solution:
Step 1: Calculate photon energy
\[
hf = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{200 \times 10^{-9}} \approx 9.94 \times 10^{-19}\, \text{J}
\]
Convert to eV:
\[
hf \approx \frac{9.94 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 6.21\, \text{eV}
\]
Step 2: Use the photoelectric equation
\[
K_{max} = hf - \phi = 6.21\, \text{eV} - 2.2\, \text{eV} = 4.01\, \text{eV}
\]
Answer:
Maximum kinetic energy is approximately 4.01 eV.
Advanced Problems and Conceptual Clarifications
While the above problems cover fundamental applications, more advanced exercises involve multiple photons, varying intensities, or energy distributions.
Problem Type 4: Effect of Intensity on Electron Kinetic Energy
Question:
Does increasing the intensity of incident light increase the maximum kinetic energy of emitted electrons? Justify your answer.
Answer:
In the photoelectric effect, the maximum kinetic energy of emitted electrons depends solely on the energy of individual photons (through the photon wavelength), not on the intensity of the light. Increasing the intensity increases the number of incident photons per unit time, thus increasing the number of emitted electrons (photoelectric current). However, it does not affect their maximum kinetic energy. Therefore, increasing intensity does not increase the maximum kinetic energy.
Problem Type 5: Multiple Photons and Nonlinear Effects
Question:
Explain what happens when the incident light has a photon energy less than the work function, but the light intensity is very high. Can electrons be emitted?
Answer:
In the classical photoelectric effect, electrons are emitted only if individual photons have enough energy to overcome the work function. Increasing the intensity (number of photons) at energies below threshold does not cause electron emission because no single photon has sufficient energy to eject an electron. Multi-photon absorption processes, where multiple photons are absorbed simultaneously, can occur at very high intensities, allowing electrons to be emitted even if individual photons are below the threshold. This nonlinear process is the basis of multi-photon photoemission phenomena.
Summary and Key Takeaways
- The photoelectric effect provides evidence for the quantization of light, with photon energy directly proportional to frequency.
- Problems often involve converting wavelength to energy, applying Einstein’s equation, and understanding the relationship between work function, photon energy, and kinetic energy.
- The threshold wavelength is inversely proportional to the work function; higher work functions correspond to shorter threshold wavelengths.
- Increasing light intensity increases the number of emitted electrons but does not affect their maximum kinetic energy.
- Multi-photon processes can enable electron emission with photon energies below the work function, but such effects require high light intensities and are nonlinear.
Conclusion
Mastery of photoelectric effect problems is essential for grasping the quantum nature of light and matter interactions. By practicing a variety of problems—ranging from straightforward calculations to conceptual questions—students and researchers can develop a robust understanding of this pivotal phenomenon. Correctly applying the principles, equations, and concepts outlined here will facilitate accurate problem-solving and deepen insight into quantum physics.
References:
- Tipler, P. A., & Llewellyn, R. (2008). Modern Physics. W. H. Freeman.
- Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics. Wiley.
- Griffiths, D. J. (2017). Introduction to Quantum Mechanics. Cambridge University Press.
Note: Always verify units and convert them appropriately to avoid common errors in calculations.
Question Answer What is the photoelectric effect and how is it related to the kinetic energy of ejected electrons? The photoelectric effect occurs when photons strike a metal surface and eject electrons. The kinetic energy of the emitted electrons is given by the equation KE = hf - φ, where hf is the photon energy and φ is the work function of the metal. This shows that the electrons' kinetic energy depends on the photon energy, not on the intensity of light. How do you calculate the work function of a metal using photoelectric effect data? The work function φ can be calculated using the threshold frequency (f₀) at which electrons just start to eject: φ = hf₀. Alternatively, using the stopping potential (V₀), the work function can be found from φ = hf₀ = eV₀ + KE_max, where KE_max is the maximum kinetic energy of the emitted electrons at the threshold. A photon of wavelength 500 nm hits a metal surface and ejects electrons with a maximum kinetic energy of 2 eV. What is the work function of the metal? First, calculate the photon energy: hf = (6.626×10⁻³⁴ Js)(3×10⁸ m/s) / 500×10⁻⁹ m ≈ 3.97×10⁻¹⁹ J ≈ 2.48 eV. Using KE_max = 2 eV, the work function φ = hf - KE_max ≈ 2.48 eV - 2 eV = 0.48 eV. If the frequency of incident light is doubled, how does it affect the kinetic energy of the emitted electrons? The kinetic energy of the emitted electrons increases linearly with the photon energy. Doubling the frequency doubles the photon energy (hf), thereby increasing the maximum kinetic energy of the electrons by an amount equal to the increase in photon energy minus the work function: KE_max = hf - φ. Therefore, the electrons will have higher kinetic energy. In a photoelectric experiment, increasing the intensity of incident light increases the number of emitted electrons but does not change their maximum kinetic energy. Why? Because the maximum kinetic energy of emitted electrons depends on the photon energy (frequency), not on the light's intensity. Increasing intensity increases the number of photons and thus the number of emitted electrons, but since photon energy remains unchanged, KE_max remains the same. A metal emits photoelectrons when illuminated with light of wavelength 400 nm. If the work function of the metal is 2 eV, what is the maximum kinetic energy of the emitted electrons? Calculate photon energy: hf = (6.626×10⁻³⁴ Js)(3×10⁸ m/s) / 400×10⁻⁹ m ≈ 4.97×10⁻¹⁹ J ≈ 3.11 eV. Subtracting the work function: KE_max = 3.11 eV - 2 eV = 1.11 eV. Therefore, the maximum kinetic energy is approximately 1.11 eV. Related keywords: photoelectric effect, photoelectric equation, work function, photon energy, Einstein's photoelectric equation, threshold frequency, incident light intensity, maximum kinetic energy, photoelectric current, example problems
