Explanation 1) Concept: By using the information given in the problem, we can find the width of the lines L x and L y . Then by substituting all the values in the formula for energy of electron in two dimension, given by equation 39 - 20 , we can find the energy of the electron. 2) Formula: E n x , n y = h 2 8 m n x 2 L x 2 + n y 2 L y 2 3) Given: i) Number of maxima along line that bisects L x is n y = 3 ii) Number of maxima along line that bisects L y is n x = 5 iii) Separation between of maxima along line that bisects L x is 2 n m iv) Separation between of maxima along line that bisects L y is 3 n m 4) Calculations: The line that bisects L x is a line parallel to y -axis and its equation is given as x = L x 2 Along this line, there are three points at which the detection probability is maximum, i.e. n y = 3 Since the maxima are separated by 2 n m , the width of the line is L y = n y 2 n m = 3 2 n m = 6 n m Similarly, the line that bisects L y is a line parallel to x -axis and its equation is given as y = L y 2 Along this line, there are five points at which the detection probability is maximum, i.e

Fundamentals of Physics Extended 10e Binder Ready Version + WileyPLUS Registration Card

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Chapter 39, Problem 24P

To determine

To find:

The energy of the electron.

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Fundamentals of Physics Extended 10e Binder Ready Version + WileyPLUS Registration Card

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