Question

Consider a three dimensional rectangular infinite potential well with sides of length L, 2L and 3L.

What is the energy of the first excited state relative to the energy of the ground state?
What is the energy of the second excited state relative to the energy of the ground state?
What is the energy of the third excited state relative to the energy of the ground state?
What is the energy of the fourth excited state relative to the energy of the ground state?

Answer 1

For one dimensional rectangular box of length,

E = k n² / L²                                 { eq. 1 }

where, k = (hbar)²² / 2m

inserting the value of k in above eq.

E = (hbar)²² n² / 2m L²       { eq.2 }

For three dimensional box,

E = k [n²/L² + m²/(2L)² + p²/(3L)²]                            where, n = 1,2,3,.. m = 1,2,3,... p = 1,2,3,...

At n = 1, m = 1 & p = 1,

on the ground state, energy is given as ::

E = k [1/ L² + 1/4 L² + 1/9 L²]    

E = k [(36 + 9 + 4) / (36 L²)]    

E = 49 k / 36 L²                                                 { eq.3 }

(a) the energy of the first excited state relative to the energy of the ground state which is given as ::

At n=m=1, p = 2, energy of first excited state -

E₁ = k [1/ L² + 1/4 L² + 4/9 L²]    

E₁ = 61 k / 36 L²                                 { eq.4 }

and   E₁ / E = (61 k / 36 L²) x (36 L² / 49 k)

E₁ / E = 61 / 49

E₁ / E = 1.24

(b) the energy of the second excited state relative to the energy of the ground state is given as ::

At n=1, m=2 , p=1

energy of the second excited state, E₂ = k [1/ L² + 4/4 L² + 1/9 L²]    

E₂ = 76 k / 36 L²                               { eq.5 }

and   E₂ / E = (76 k / 36 L²) x (36 L² / 49 k)

E₂ / E = 76 / 49

E₂ / E = 1.55

(c) the energy of the third excited state relative to the energy of the ground state is given as :

At n=1, m=1 , p=3

energy of the second excited state, E₃ = k [1/ L² + 1/4 L² + 9/9 L²]    

E₃ = 81 k / 36 L²                                               { eq.6 }

and   E₃ / E = (81 k / 36 L²) x (36 L² / 49 k)

E₃ / E = 81 / 49

E₃ / E = 1.65

(d) the energy of the fourth excited state relative to the energy of the ground state which is given as :

At n=1, m=2 , p=2

energy of the second excited state, E₄ = k [1/ L² + 4/4 L² + 4/9 L²]    

E₄ = 88 k / 36 L²                                                { eq.7 }

and   E₄ / E = (88 k / 36 L²) x (36 L² / 49 k)

E₄ / E = 88 / 49

E₄ / E = 1.79