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Chapter 25: Problem 15

If \(K_{1}\) and \(K_{2}\) are the maximum kinetic energies of photoelectrons emitted when lights of wavelength \(\lambda_{1}\) and \(\lambda_{2}\) respectively incident on a metallic surface and \(\lambda_{1}=3 \lambda_{2}\), then (a) \(K_{1}>\frac{K_{2}}{3}\) (b) \(K_{1}<\frac{K_{2}}{3}\) (c) \(K_{1}=3 K_{2}\) (d) \(K_{2}=3 K_{1}\)

Short Answer

Expert verified
Option (a) \( K_1 > \frac{K_2}{3} \) is correct.

Step by step solution

01

Understand the Photoelectric Effect Equation

The energy of incident light and the work function governs the kinetic energy of emitted photoelectrons. The equation is given by: \[ K = hf - ext{Work function} \] where \( K \) is the kinetic energy of the emitted photoelectrons, \( h \) is Planck's constant, and \( f \) is the frequency of the light.
02

Relate Frequency and Wavelength

The frequency \( f \) is related to the wavelength \( \lambda \) of light by the equation \( c = \lambda f \), where \( c \) is the speed of light. Therefore, \( f = \frac{c}{\lambda} \). Using this relationship, the kinetic energy equation becomes: \[ K = \frac{hc}{\lambda} - \text{Work function} \]
03

Establish Kinetic Energy Relationships

Given \( \lambda_1 = 3\lambda_2 \), the relation for kinetic energies will be: \[ K_1 = \frac{hc}{\lambda_1} - \text{Work function} \] \[ K_2 = \frac{hc}{\lambda_2} - \text{Work function} \]. Therefore, \[ K_1 = \frac{hc}{3\lambda_2} - \text{Work function} \]
04

Analyze Expression for K_1 and K_2

Using the derived expressions for kinetic energies: \[ K_1 = \frac{1}{3} \left( \frac{hc}{\lambda_2} \right) - \text{Work function} \] and \[ K_2 = \frac{hc}{\lambda_2} - \text{Work function} \]. Compare \( K_1 \) and \( \frac{K_2}{3} \): \[ \frac{K_2}{3} = \frac{1}{3} \left( \frac{hc}{\lambda_2} - \text{Work function} \right) \]
05

Compare K1 and K2/3

Substituting and comparing: \[ K_1 = \frac{1}{3} \frac{hc}{\lambda_2} - \text{Work function} \] and \[ \frac{K_2}{3} = \frac{1}{3} \left( \frac{hc}{\lambda_2} - \text{Work function} \right) \]. Since the work function affects both energies equally, and \( \frac{1}{3} \frac{hc}{\lambda_2} > \frac{1}{3} \left( \frac{hc}{\lambda_2} - \text{Work function} \right) \), it follows that \( K_1 > \frac{K_2}{3} \).
06

Conclusion

Based on the comparison in Step 5, option (a) \( K_1 > \frac{K_2}{3} \) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of the photoelectric effect, this concept specifically refers to the energy of photoelectrons emitted from a metallic surface when exposed to light.
  • The energy of light photons must be greater than or equal to the work function of the metal to release electrons.
  • In the equation of the photoelectric effect, kinetic energy is expressed as \( K = hf - \text{Work function} \).
  • Here, \( K \) represents the kinetic energy of the photoelectrons leaving the metal.
Understanding kinetic energy in the context of photoelectricity is crucial, as it helps explain how light interacts with matter to release electrons.
Wavelength
Wavelength is the distance between consecutive crests of a wave, typically measured in meters. It plays a significant role in determining the frequency and energy of light.
  • The relationship between wavelength \( \lambda \) and frequency \( f \) is given by the equation \( c = \lambda f \), where \( c \) is the speed of light.
  • In the photoelectric effect, the wavelength of the incident light determines its energy, which is crucial for discussing photoelectric emission.
  • Shorter wavelengths mean higher frequencies, thus higher energy photons are more capable of overcoming the work function of a metal.
This is why experimentation with different wavelengths helps us understand electron emissions from surfaces.
Planck's Constant
Planck's constant \( h \) is a fundamental constant that is vital in quantum mechanics. It plays a key role in the photoelectric effect's equation.
  • Planck's constant has a value of approximately \( 6.626 \times 10^{-34} \) Js.
  • The equation \( E = hf \) shows Planck's constant as a proportional factor between energy \( E \) of a photon and its frequency \( f \).
  • It's central to understanding how energy levels and quantum states relate in quantum physics.
Through the concept of Planck's constant, we delve deeper into the quantum nature of light and energy, enriching our grasp of the microscopic world.
Frequency
Frequency refers to how many wave cycles pass a point per unit of time. In terms of light, it's an essential factor in determining energy and behavior in the photoelectric effect.
  • It's commonly measured in Hertz (Hz), where one hertz equals one cycle per second.
  • Frequency \( f \) is inversely proportional to wavelength \( \lambda \), meaning \( f = \frac{c}{\lambda} \).
  • Higher frequency means higher energy, as indicated by the equation \( E = hf \).
In the photoelectric effect, adjusting light frequencies helps demonstrate how energy interacts with surfaces, allowing electrons to escape.
Work Function
The work function is the minimum energy required to eject an electron from the surface of a material. This property is crucial in the study of photoelectric effect.
  • Each element has a specific work function, usually measured in electron volts (eV).
  • The concept is crucial because, for an electron to be released, the incoming photon energy must exceed this threshold.
  • If the photon's energy is lower than the work function, no electrons will be emitted no matter how intense the light.
The work function helps in determining the threshold frequency and wavelength necessary for photoemission, deepening our understanding of how light and matter interact.

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Most popular questions from this chapter

In the photoelectric effect, electrons are emitted (a) at a rate that is independent of the emitter (b) at a rate that is proportional to the square of the amplitude of the incident radiation (c) with a maximum velocity proportional to the frequency of incident radiation (d) only if the frequency of the incident radiation is above a threshold value

The energy of a photon of characteristic \(X\) -rays from a Coolidge tube comes from the (a) kinetic energy of the striking electron (b) kinetic energy of the free electron of target (c) kinetic energy of the ions of the target (d) electronic transition of the target atom

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is \(E\). Let \(\lambda_{1}\) be the de Broglie wavelength of the proton and \(\lambda_{2}\) be the wavelength of the photon. The ratio \(\lambda_{1} / \lambda_{2}\) is proportional to (a) \(E^{o}\) (b) \(E^{1 / 2}\) (c) \(E^{-1}\) (d) \(E^{-2}\)

Which of the following statements is false? (a) Material wave (de Broglie wave) can travel in vacuum (b) Electromagnetic wave can travel through vacuum (c) The velocity of photon is not the same whether light passes through any medium or vacuum (d) Wavelength of de Broglie wave depends upon velocity

The characteristic of the electrons striking the target in an \(X\) -ray tube that determines the intensity of \(X\) -rays is (a) energy (b) momentum (c) number incident per second (d) mass
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