consider an electron freely moving ina two-dimensional square boxof length 2.50

Question

consider an electron freely moving ina two-dimensional square boxof length 2.50 nm. a) evaluae the zero-point energy of the particle. b) what is the origin of this zero-point energy? c) now consider anoother state of the electron that has energytwenty five times the zero-point energy. what is the degeneracy ofthis energy level? a) evaluae the zero-point energy of the particle. b) what is the origin of this zero-point energy? c) now consider anoother state of the electron that has energytwenty five times the zero-point energy. what is the degeneracy ofthis energy level?

Explanation / Answer

For a two dimensional box, the general expression forenergy is Enx,ny = (h2 /8m) {(n2x / Lx)+(n2y / Ly)} For a square box, Lx =Ly = Lso we have Enx,ny = (h2 /8mL) {n2x   + n2y} -------------------- Zero point energy is nothin but lowest energy, for whichnx = ny = 1 So zero point energy is E1,1  = ((6.625*10^-34J.s)2 /8(9.19*10^-31kg)*2.50nm(10^-9m/nm)){12x   + 12y} ----------------------- The particle (electron) is not static, it is always in motionso that it has some energy. Which is in accord to Heisen berguncertainity principle. ------------------ E1,1 = 2 (h2 / 8mL) is zero pointenergy. Twotenty five times to zero point energy is 25E1,1 =(25*2)(h2 / 8mL)                                                                             = 50 (h2 / 8mL) So n2x   + n2y = 50 To know the number of pairs, satisfying the abovecondition. 1^2 +7^2 = 50     5^2 +5^2 = 50 4^2 + 6^2 = 50 So we have three pairs; (1,7) , (5,5) , (4,6) ;Hence degenercy is 3 E1,1  = ((6.625*10^-34J.s)2 /8(9.19*10^-31kg)*2.50nm(10^-9m/nm)){12x   + 12y} ----------------------- The particle (electron) is not static, it is always in motionso that it has some energy. Which is in accord to Heisen berguncertainity principle. ------------------ E1,1 = 2 (h2 / 8mL) is zero pointenergy. Twotenty five times to zero point energy is 25E1,1 =(25*2)(h2 / 8mL)                                                                             = 50 (h2 / 8mL) So n2x   + n2y = 50 To know the number of pairs, satisfying the abovecondition. 1^2 +7^2 = 50     5^2 +5^2 = 50 4^2 + 6^2 = 50 So we have three pairs; (1,7) , (5,5) , (4,6) ;Hence degenercy is 3

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