Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A light detector has an absorbing area of $2.00 \times 10^{- 6} m^{2}$ and absorbs 50% of the incident light, which is at a wavelength 600nm. The detector faces an isotropic source, 12.0 m from the source. The energy E emitted by the source versus time t is given in Fig. 38-26 ( $E_{s} = 7.2 n J$ , $t_{s} = 2.0 s$ ). At what rate are photons absorbed by the detector?

The rate of absorption of the photons by the detector is 6.0 photons/s.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Describe the expression of emission rate and energy of the photon:

The expression of emission rate is given by,

$R = \frac{P}{E_{p}}$

Here, P is power.

The expression for the frequency of the photon is given by,

$f = \frac{c}{λ}$

Here, c is the speed of light and $λ$ is the wavelength.

The energy of a photon is given by,

E=hf

Here, h is the Planck’s constant.

Combine the above two equations.

$E = \frac{h c}{λ}$ ….. (1)

Determine the rate of absorption of the photons by detector:

Consider the given data as below.

The wavelength, $λ = 600 nm$

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

The speed of light, $c = 3 \times 10^{8} m / s$

The time, $Δ t = 2.0 s$

The energy, $Δ E = 7.2 nJ$

Substitute the below values in eq 1.

The wavelength, $λ = 600 nm$

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

The speed of light, $c = 3 \times 10^{8} m / s$

$E = \frac{(6.626 \times 10^{- 34} J \cdot s) (3 \times 10^{8} m / s)}{(600 nm) (\frac{1 m}{10^{9}})} = (3.3125 \times 10^{- 19} J) (\frac{1 eV}{1.6 \times 10^{- 19} J}) = 2.07 eV$

The expression of average power output of the source is given by,

$P_{emit} = \frac{Δ E}{Δ t}$ ….. (2)

Substitute 7.2 nJ for $Δ E$ , and 2.0 s for $Δ t$ in equation (2).

$P_{emit} = \frac{7.2 nJ}{2.0 s} = \frac{(7.2 nJ) (\frac{1.0 e V}{1.6 \times 10^{- 19}})}{2.0 s} = = 2.25 \times 10^{10} e V / s$

Define the rate at which photons radiate.

$R_{emit} = \frac{P_{emit}}{E} = \frac{2.25 \times 10^{10} e V / s}{2.07 eV} = 1.09 \times 10^{10} p h o t o n s / s$

Since the source is isotropic, then the flux away is just the power radiated divided by the surface area of the sphere.

$\begin{matrix} I = \frac{P}{A} \\ = \frac{P_{emit}}{4 π r^{2}} \end{matrix}$

If the detector absorbs light at 50% efficiency, then the power is,

$P_{abs} = (\frac{50}{100}) I A_{detect} R_{abs} E = (0.5) (\frac{P_{emit}}{4 {πr}^{2}}) A_{detect} R_{abs} = (0.5) (\frac{A_{detect}}{4 {πr}^{2}}) (\frac{P_{emit}}{E})$ ….. (3)

Substitute $2.0 \times 10^{- 6} m^{2}$ for $A_{detect},$ 12.0m for r, and $1.09 \times 10^{10} p h o t o n s / s$ for $R_{emit}$ in equation (3).

Hence, the rate of absorption of the photons by detector is 6.0 photons/s.

	Electron Energy	Barrier Height	Barrier Thickness
(a)	E	5E	L
(b)	E	17E	L/2
(c)	E	2E	2L

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