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or a quantum particle confined inside a cubic box of side L, the ground state energy is given by E0.  The energy of the first excited state is
  • a)
    2E0
  • b)
    2√E0
  • c)
    3E0
  • d)
    6E0
Correct answer is option 'A'. Can you explain this answer?
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or a quantum particle confined inside a cubic box of side L, the groun...
E0 is correct.

For a particle confined inside a cubic box of side L, the energy levels are given by:

E(n1, n2, n3) = [(n1^2 + n2^2 + n3^2) * h^2] / [8mL^2]

where n1, n2, and n3 are integers that specify the quantum numbers for the three dimensions, h is the Planck constant, and m is the mass of the particle.

The ground state corresponds to n1 = n2 = n3 = 1, and thus has energy:

E0 = [(1^2 + 1^2 + 1^2) * h^2] / [8mL^2] = (3/8) * (h^2 / mL^2)

The first excited state corresponds to one of the quantum numbers being 2, and the other two being 1. Without loss of generality, let n1 = 2, and n2 = n3 = 1. Then the energy of this state is:

E1 = [(2^2 + 1^2 + 1^2) * h^2] / [8mL^2] = (6/8) * (h^2 / mL^2) = (3/4) * E0

Therefore, the correct answer is (b) 2E0.
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or a quantum particle confined inside a cubic box of side L, the ground state energy is given by E0. The energy of the first excited state isa)2E0b)2√E0c)3E0d)6E0Correct answer is option 'A'. Can you explain this answer?
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