Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Suppose the fractional efficiency of a Cesium surface (with work function 1.80 eV ) is $1.0 \times 10^{- 16}$ ; that is, on average one electron is ejected for every $10^{16}$ photons that reach the surface. What would be the current of electrons ejected from such a surface if it were illuminated with 600 nm light from 2.0 mW laser and all the ejected electrons took part in the charge flow?

The required current of electrons ejected from a surface is $9.68 \times 10^{- 20} A .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Identification of the given data:

The given data is listed below.

The fractional efficiency is $1.0 \times 10^{- 16}$

The wavelength of the laser is $λ = 600 n m .$

Power, P= 2.0 mW.

The laser output:

The output of the laser is given by

P=RE

Where R is the rate of the number of photons emitted per unit time of the laser and E is the energy of a single photon.

To determine the current of electrons ejected from the surface

The power output is given by

P = RE ….. (1)

Now, Energy is given by,

$E = \frac{h c}{λ}$

Here, h is the Plank’s constant, c is the speed of light and $λ$ is the wavelength.

Substitute localid="1663150706210" $E = \frac{h c}{λ}$ for E into equation (1)

localid="1663150711000" $P = \frac{h c R}{λ}$

Solving the above equation and finding the value of R.

localid="1663150715861" $R = \frac{P λ}{h c}$

Consider the given data below.

The wavelength, localid="1663150719899" $λ = 600 n m$

The power, localid="1663150723665" $P = 2.0 m W$

Plank’s constant, localid="1663150727678" $h = 6.626 \times 10^{- 34} J \cdot s$

Speed of light, localid="1663150733098" $c = 2.998 \times 10^{8} m / s$

Substitute these numerical values

Now, the fractional efficiency of Cesium surface is $1.0 \times 10^{- 16} .$

Therefore, the rate of photons that actually cause photoelectric emissions is

$R' = 1.0 \times 10^{- 16} R = (1.0 \times 10^{- 16}) (6.04 \times 10^{15} photons / s) = 0.604 electron / s .$

Now, the current is equal to the rate of electrons multiplied by the charge of an electron

$i = R' e = (0.604 s^{- 1}) (1.602 \times 10^{- 19} p h o t o n s / s) = 9.68 \times 10^{- 20} A$

Hence, the current of electrons ejected from the surface is $9.68 \times 10^{- 20} A .$

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Identification of the given data:

The laser output:

To determine the current of electrons ejected from the surface

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Identification of the given data:

The laser output:

To determine the current of electrons ejected from the surface

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics