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The two energy levels (nx = 1, ny = 6) and (nx = 3,ny = 2) of a particle in a two-dimensional rectangular box (potential is zero inside, and infinite outside) of sides Lx and Ly are found to be degenerate. If Lx = 1 in appropriate units, then Ly is
- a)2
- b)3
- c)4
- d)6
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The two energy levels (nx= 1, ny= 6) and (nx= 3,ny= 2) of a particle i...
- When nx = 1, ny = 6, Lx = 1 and Ly= ly, the energy will be as follows,
- Also, when nx = 3, ny = 2, Lx = 1 and Ly= ly, the energy will be as follows,
- As the energy levels are degenerate (E16=E32), thus we got,
or, ly2 = 4
or, ly = 2 - Thus, If Lx = 1 in appropriate units, then Ly is 2.
Hence, Ly is 2.
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The two energy levels (nx= 1, ny= 6) and (nx= 3,ny= 2) of a particle i...
The given problem involves a particle in a two-dimensional rectangular box with sides Lx and Ly. The potential inside the box is zero, while it is infinite outside. The question asks us to determine the value of Ly if the energy levels (nx= 1, ny= 6) and (nx= 3, ny= 2) are degenerate.
Understanding Degeneracy:
Degeneracy refers to the phenomenon where different quantum states have the same energy. In this case, if two different energy levels are degenerate, it means that the particle can have the same energy in both states, despite having different quantum numbers.
Determining the Energy Levels:
The energy levels of a particle in a rectangular box are given by the equation:
E = (nx^2 * π^2 * ℏ^2) / (2mLx^2) + (ny^2 * π^2 * ℏ^2) / (2mLy^2)
where nx and ny are the quantum numbers in the x and y directions, respectively, ℏ is the reduced Planck's constant, m is the mass of the particle, and Lx and Ly are the lengths of the box in the x and y directions, respectively.
Since the problem states that the energy levels (nx= 1, ny= 6) and (nx= 3, ny= 2) are degenerate, we can equate their energies:
(nx=1, ny=6): E1 = (π^2 * ℏ^2) / (2mLx^2) + (36 * π^2 * ℏ^2) / (2mLy^2)
(nx=3, ny=2): E2 = (9 * π^2 * ℏ^2) / (2mLx^2) + (4 * π^2 * ℏ^2) / (2mLy^2)
Setting E1 = E2, we can solve for Ly:
(π^2 * ℏ^2) / (2mLx^2) + (36 * π^2 * ℏ^2) / (2mLy^2) = (9 * π^2 * ℏ^2) / (2mLx^2) + (4 * π^2 * ℏ^2) / (2mLy^2)
Simplifying the equation:
(36 * π^2 * ℏ^2) / (2mLy^2) - (4 * π^2 * ℏ^2) / (2mLy^2) = (9 * π^2 * ℏ^2) / (2mLx^2) - (π^2 * ℏ^2) / (2mLx^2)
(32 * π^2 * ℏ^2) / (2mLy^2) = (8 * π^2 * ℏ^2) / (2mLx^2)
Canceling out common terms and simplifying further:
Ly^2 = 4Lx^2
Since Lx = 1, substituting this value:
Ly^2 = 4
Taking the square root of both sides, we get:
Ly = 2
Therefore, the correct answer is option 'A' - Ly = 2.
Understanding Degeneracy:
Degeneracy refers to the phenomenon where different quantum states have the same energy. In this case, if two different energy levels are degenerate, it means that the particle can have the same energy in both states, despite having different quantum numbers.
Determining the Energy Levels:
The energy levels of a particle in a rectangular box are given by the equation:
E = (nx^2 * π^2 * ℏ^2) / (2mLx^2) + (ny^2 * π^2 * ℏ^2) / (2mLy^2)
where nx and ny are the quantum numbers in the x and y directions, respectively, ℏ is the reduced Planck's constant, m is the mass of the particle, and Lx and Ly are the lengths of the box in the x and y directions, respectively.
Since the problem states that the energy levels (nx= 1, ny= 6) and (nx= 3, ny= 2) are degenerate, we can equate their energies:
(nx=1, ny=6): E1 = (π^2 * ℏ^2) / (2mLx^2) + (36 * π^2 * ℏ^2) / (2mLy^2)
(nx=3, ny=2): E2 = (9 * π^2 * ℏ^2) / (2mLx^2) + (4 * π^2 * ℏ^2) / (2mLy^2)
Setting E1 = E2, we can solve for Ly:
(π^2 * ℏ^2) / (2mLx^2) + (36 * π^2 * ℏ^2) / (2mLy^2) = (9 * π^2 * ℏ^2) / (2mLx^2) + (4 * π^2 * ℏ^2) / (2mLy^2)
Simplifying the equation:
(36 * π^2 * ℏ^2) / (2mLy^2) - (4 * π^2 * ℏ^2) / (2mLy^2) = (9 * π^2 * ℏ^2) / (2mLx^2) - (π^2 * ℏ^2) / (2mLx^2)
(32 * π^2 * ℏ^2) / (2mLy^2) = (8 * π^2 * ℏ^2) / (2mLx^2)
Canceling out common terms and simplifying further:
Ly^2 = 4Lx^2
Since Lx = 1, substituting this value:
Ly^2 = 4
Taking the square root of both sides, we get:
Ly = 2
Therefore, the correct answer is option 'A' - Ly = 2.
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The two energy levels (nx= 1, ny= 6) and (nx= 3,ny= 2) of a particle in a two-dimensional rectangular box (potential is zero inside, and infinite outside) of sides Lxand Lyare found to be degenerate. If Lx= 1 in appropriate units, then Lyisa)2b)3c)4d)6Correct answer is option 'A'. Can you explain this answer?
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