Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Figure 39-30 shows a two-dimensional, infinite-potential well lying in an xy plane that contains an electron. We probe for the electron along a line that bisects $L_{x}$ and find three points at which the detection probability is maximum. Those points are separated by 2.00 nm . Then we probe along a line that bisects $L_{y}$ and find five points at which the detection probability is maximum. Those points are separated by 3.00 nm . What is the energy of the electron?

The energy of the electron is 0.136 eV.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The energy of two-dimensional electron traps:

The quantized energies for an electron trapped in a two-dimensional infinite potential well that forms a rectangular corral are,

$E = \frac{h^{2}}{8 m} (\frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}})$ ….. (1)

Here, $n_{x}$ is quantum number for which the electron’s matter wave fits in well width $L_{x}$ , $n_{y}$ is quantum number for which the electron’s matter wave fits in well width $L_{y}$ , h is plank constant, and is mass of the electron.

Step 2: Find the energy of the electron:

Every probability maximum represents a quantum number, in this case in x- direction, $n_{x} = 3$ and in y-direction $n_{y} = 5$ .

Substitute localid="1661774839620" $6.626 \times 10^{- 34} J \cdot s for h, 9.109 \times 10^{- 31} kg for m, 3 for n_{x}, 5 for n_{y}, 2 \times 10^{- 9} m for L_{x} . and 3 \times 10^{- 9} m for L_{y}$ in equation (1).

$E = \frac{{(6.626 \times 10^{- 34} J . s)}^{2}}{(8) (9.109 \times 10^{- 31} kg)} (\frac{3^{3}}{{(3.2 \times 10^{- 9} m)}^{2}} + \frac{5^{2}}{{(5.3 \times 10^{- 9} m)}^{2}}) = 2.18 \times 10^{- 20} J = (2.18 \times 10^{- 20} J) (\frac{6.242 \times 10^{18} eV}{1 J}) = 0.136 eV$

Hence, the energy of the electron is 0.136 eV .

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