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Chapter 25: Problem 24

The work function of a metal is \(1 \mathrm{eV}\). Light of wavelength \(3000 \AA\) is incident on this metal surface. The velocity of emitted photoelectrons will be (a) \(10 \mathrm{~ms}^{-1}\) (b) \(10^{3} \mathrm{~ms}^{-1}\) (c) \(10^{4} \mathrm{~ms}^{-1}\) (d) \(10^{6} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity of the emitted photoelectrons is (d) 10^6 \text{ m/s}.

Step by step solution

01

Calculate Energy of Incident Light

First, we need to determine the energy of the incoming light using its wavelength. The formula to convert wavelength \(\lambda\) to energy \(E\) is given by:\[ E = \frac{hc}{\lambda} \]where \( h \) is the Planck's constant \( 4.1357 \times 10^{-15} \text{ eV} \cdot \text{s} \) and \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). The given wavelength is \(3000\, \text{脜} = 3000 \times 10^{-10} \text{ m} \). Substituting the values gives:\[ E = \frac{4.1357 \times 10^{-15} \times 3 \times 10^8}{3000 \times 10^{-10}} \approx 4.14 \text{ eV} \]
02

Apply Photoelectric Equation

The energy of incident light \(4.14 \text{ eV}\) is used to liberate an electron from the metal, overcoming the work function (given) and providing kinetic energy to the electron. Using the photoelectric equation:\[ KE_{max} = E - \phi \]where \( \phi \) (work function) is \(1 \text{ eV}\). Thus:\[ KE_{max} = 4.14 - 1 = 3.14 \text{ eV} \]
03

Convert Kinetic Energy to Velocity

Convert the kinetic energy obtained into velocity. Kinetic energy for electrons is given by:\[ KE_{max} = \frac{1}{2}mv^2 \]Rearranging and solving for \(v\) gives:\[ v = \sqrt{\frac{2 \times KE_{max}}{m}} \]where the mass \(m\) of an electron is \(9.11 \times 10^{-31} \text{ kg}\). Converting \(KE_{max}\) to joules (since 1 eV = \(1.6 \times 10^{-19} \text{ J}\)):\[ KE_{max} = 3.14 \times 1.6 \times 10^{-19} \text{ J} \approx 5.024 \times 10^{-19} \text{ J} \]Now substitute these into the velocity equation:\[ v = \sqrt{\frac{2 \times 5.024 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 10^6 \text{ m/s} \]
04

Match Result with Given Options

The calculated velocity of the emitted photoelectrons is \(10^6 \text{ m/s}\). Reviewing the options given:- (a) \(10 \text{ m/s}\)- (b) \(10^3 \text{ m/s}\)- (c) \(10^4 \text{ m/s}\)- (d) \(10^6 \text{ m/s}\)The correct option matches with (d) \(10^6 \text{ m/s}\).

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Key Concepts

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

Work Function
The work function is a key concept in the photoelectric effect. It represents the minimum amount of energy needed to eject an electron from the surface of a metal. Each metal has a unique work function, denoted by \( \phi \), and is typically measured in electron volts (eV).
  • The work function must be overcome by the incident photon for an electron to be emitted from the metal.
  • In the given exercise, the work function is \(1 \text{ eV}\).
  • When photons with energy above this threshold strike the metal, they provide the electron with enough energy to break free, provided by the energy surplus of the photon.
Understanding the work function is crucial to grasping how different metals react to different wavelengths and intensities of light.
This concept explains why some metals are more reactive to light than others in photoelectric applications.
Kinetic Energy
Kinetic energy in the photoelectric effect represents the energy that is transferred to an electron once it has overcome the work function of the metal. The kinetic energy of the emitted electron is calculated by subtracting the work function from the photon energy.
  • In the photoelectric equation: \( KE_{\text{max}} = E - \phi \), where \(E\) is the photon energy and \(\phi\) is the work function.
  • For the exercise, \( KE_{\text{max}} = 4.14 \text{ eV} - 1 \text{ eV} = 3.14 \text{ eV}\).
  • This value indicates how much translational energy an electron gains post-ejection.
Once the kinetic energy is known, it can be converted to other forms such as velocity, demonstrating how fast the electron will move in the space outside the metal surface.
This kinetic movement is fundamental to understanding phenomena like electricity and conductivity in photonic and electronic devices.
Photon Energy
Photon energy is a critical concept in studying the photoelectric effect. It represents the energy carried by a photon, which is dependent on its wavelength or frequency. The formula \( E = \frac{hc}{\lambda} \) connects photon energy to its wavelength, where \( h \) is Planck's constant and \( c \) is the speed of light.
  • Longer wavelengths correspond to lower energies, while shorter wavelengths have higher energies.
  • In this exercise, photons of wavelength \(3000 \text{ 脜}\) have an energy of approximately \(4.14 \text{ eV}\).
  • This value surpasses the work function \(1 \text{ eV}\), enabling electrons to be ejected with additional energy translated into kinetic energy.
Photon energy is vital as it determines whether a specific wavelength can initiate the photoelectric effect for a given metal.
This understanding is used in various technologies, including solar cells and photo detectors, where tuning the wavelength can optimize performance.

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Most popular questions from this chapter

A parallel beam of light is incident normally on a plane surface absorbing \(40 \%\) of the light and reflecting the rest. If the incident beam carries \(60 \mathrm{~W}\) of power, the force exerted by it on the surface is (a) \(3.2 \times 10^{-8} \mathrm{~N}\) (b) \(3.2 \times 10^{-7} \mathrm{~N}\) (c) \(5.12 \times 10^{-7} \mathrm{~N}\) (d) \(5.12 \times 10^{-8} \mathrm{~N}\)

Which one of the following statements regarding photo-emission of electrons is correct? (a) Kinetic energy of electrons increases with the intensity of incident light (b) Electrons are emitted when the wavelength of the incident light is above a certain threshold wavelength (c) Photoelectric emission is instantaneous with the incidence of light (d) Photo electrons are emitted whenever a gas is irradiated with ultraviolet light

A photon of energy \(E\) ejects a photoelectrons from a metal surface whose work function is \(W_{0}\). If this electron enters into a uniform magnetic field of induction \(B\) in a direction perpendicular to the field and describes a circular path of radius \(r\), then the radius, \(r\) is given by (a) \(\sqrt{\frac{2 m\left(W_{0}-E\right)}{e B}}\) (b) \(\sqrt{\frac{2 e\left(E-W_{0}\right)}{m B}}\) (c) \(\sqrt{\frac{2 m\left(E-W_{0}\right)}{e B}}\) (d) \(\sqrt{\frac{2 m W_{0}}{e B}}\)

A \(100 \mathrm{~W}\) light bulb is placed at the centre of a spherical chamber of radius \(0.10 \mathrm{~m} .\) Assume that \(66 \%\) of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is (a) \(0.87 \times 10^{-6} \mathrm{~Pa}\) (b) \(1.77 \times 10^{-6} \mathrm{~Pa}\) (c) \(3.50 \times 10^{-6} \mathrm{~Pa}\) (d) None of the above

An electron is moving with an initial velocity \(\mathbf{v}=v_{0} \hat{\mathbf{i}}\) and is in a magnetic field \(\mathbf{B}=B_{0} \hat{\mathbf{j}}\).Then it's de-Broglie wavelength \(\quad\) (a) remains constant (b) increases with time (c) decreases with time (d) increases and decreases periodically
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