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Fundamentals Of Physics
Found in: Page 1215

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Short Answer

An electron is trapped in a one-dimensional infinite potential well. For what (a) higher quantum number and (b) lower quantum number is the corresponding energy difference equal to the energy difference E43 between the levels n = 4 and n = 3 ? (c) Show that no pair of adjacent levels has an energy difference equal to 2E43 .

  1. The highest quantum number is 7 .
  2. The lowest quantum number is 1 .
  3. It is not possible that adjacent levels have an energy difference of 2E43.
See the step by step solution

Step by Step Solution

Step 1: Introduction:

An electron is a negatively charged subatomic particle. It can be either free (not attached to any atom), or bound to the nucleus of an atom. Electrons in atoms exist in spherical shells of various radii, representing energy levels. The larger the spherical shell, the higher the energy contained in the electron.

Step 2: (a) Find the higher quantum number:

Assume that the quantum numbers of the pairs be n and n + 1 .

The lowest quantum number is n that is, n = 3

And, the highest quantum umber is n + 1 . That is,

n + 1 = 4

Now, calculate the lowest quantum number as follow:

For the lowest quantum number, subtract the higher energy from the lowest as follows:

En+1-En=E43 =E4-E3 =n42-n32π2h22mτ =42-32π2h22mτ =7π2h22mτ

Also,

En1+n2=n12-n22π2h22mτ2 =E =7π2h22mτ2

Therefore,

n12-n22=7

Use the following formula to solve further,

a2-b2=a+ ba-b

So, you can write,

n12-n22=7(n1-n2n1+n2=7

Thus the different factors that can be made are as follow;

n1-n2=1n1+n2=7

Hence the highest quantum number is 7 .

Step 3: (b) Find the lower quantum number:

The lower quantum number is,

n1-n2=1

Hence the lowest quantum number is 1 .

Step 4: (c) Find the adjacent levels has an energy:

n1To find n1 and n2 such that,

En1+n243=h2π22mτ2n12-n22 =2E43

Therefore,

h2π22mτ2n12-n22=2E43 …… (1)

Now substitute 4 for n1 and 3 for n2 in the above equation.

h2π22mτ242-32=2E43h2π22mτ216-9=2E437h2π22mτ2=2E43

Therefore,

2E43=27h2π22mτ2 =14h2π22mτ2

Thus the equation h2π22mτ2n12-n22=2E43

Implies,

n12-n22=14a2-b2=a+ba-b,the equationn12-n22=14 is rewritten as follows:n1-n2n1+n2=14

There are two ways in which 14 can be written one is 2×7 and the another is 14×1 .

Therefore,

(n1-n2)n1+n2=2×7(14×1)

Thus, the different factor that can be made are as follows:

n1-n2=2 ….. (2)

n1+n2=7 ….. (3)

Solving the equation (2) and (3) for n1 as follows:

2n1=9

n1=92 =4.5 =not an integer

Other factors are,

n1-n2=1 ….. (4)

n1+n2=14 ….. (5)

Solving the equation (5) and (5) for as follows:

2n1=15

n1=152 =7.5 =not an integer

Hence, it is not possible that adjacent levels having an energy difference of 2E43 .

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