Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

one-dimensional infinite well of length 200 pm contains an electron in its third excited state. We position an electron detector probe of width 2.00 pm so that it is centred on a point of maximum probability density. (a) What is the probability of detection by the probe? (b) If we insert the probe as described 1000 times, how many times should we expect the electron to materialize on the end of the probe (and thus be detected)?

(a) The probability of detection is 0.020.

(b) 20 times should we expect the electron to materialize on the end of

the probe.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Introduction

As the gas decays and gives off electrons, the detector uses a magnet to trap them in a magnetic bottle. A radio antenna then picks up very weak signals emitted by the electrons, which can be used to map the electrons' precise activity over several milliseconds.

Step 2: Concept

Probability of defection in width centred on position is given by

$p (x) = ψ_{n}^{2} (x) d x$

Probability density $ψ {}_{n}^{2}{(x)}$ for the trapped electron is

$ψ_{n}^{2} (x) = A^{2} s i n^{2} (\frac{n π}{L} x), [f o r n = 1, 2, 3, . . .]$

Step: 3 (a) Determine the probability of detection by the probe

Centred of the well, $x = \frac{L}{2}$

$\begin{array}{l} = \frac{200 p m}{2} \\ = 100 p m \end{array}$

Probability of detection at x ,

$p (x) = ψ_{n}^{2} (x) d x = {[\sqrt{\frac{2}{L} \sin (\frac{n π}{L} x)}]}^{2} d x = \frac{2}{L} \sin^{2} (\frac{n π}{L} x) d x$

Here,

n=3

Length, $L = 200 p m$

Width of the probe, dx=2.00 pm

Probability

$p (x = \frac{L}{2}) = \frac{2}{L} \sin^{2} (\frac{nπ}{L} x) dx = \frac{2}{L} \sin^{2} [(\frac{3 π}{L}) (\frac{L}{2})] dx = \frac{2}{L} \sin^{2} (\frac{3 π}{2}) dx = \frac{2}{L} dx = \frac{2}{200 pm} (2.00 pm) = 0.020$

Hence, the probability of detection is 0.020.

Step 4: (b) Determine the electron to materialize on the end of the probe

Number of independent insertion,

$N = 1000$

Number of times the electron to be detected, $n = N p$

$= (1000) (0.020) = 20$

Hence, 20 times should we expect the electron to materialize on the end of

the probe.

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Short Answer

Step by Step Solution

Step 1: Introduction

Step 2: Concept

Step: 3 (a) Determine the probability of detection by the probe

Step 4: (b) Determine the electron to materialize on the end of the probe

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