contestada

A particle of mass m is placed in a three-dimensional rect- angular box with edge lengths 2L, L, and L. Inside the box the potential energy is zero, and outside it is in nite; therefore, the wave function goes smoothly to zero at the sides of the box. Calculate the energies and give the quan- tum numbers of the ground state and the rst ve excited states (or sets of states of equal energy) for the particle in the box.

Respuesta :

Answer:

the energy of groud state = [tex]\frac{9h^{2} }{32ml^{2} }[/tex]

Explanation:

the energy of a 3 dimensional rectangular box is given by [tex]\frac{h^{2} }{8m}  (\frac{n_{x} ^{2} }{l_{x} ^{2}  }  +\frac{n_{y}^{2}  }{l_{y} ^{2} }+ \frac{n_{z}^{2}  }{l_{z} ^{2} })[/tex]

where h is planks constant m is the mass of the particle [tex]n_{x}[/tex],[tex]n_{y}[/tex] and [tex]n_{z}[/tex] are principal quantum number in x y and z direction. and [tex]l_{x}[/tex],[tex]l_{y}[/tex] and [tex]l_{z}[/tex] are length of box in x y and z direction.

therefore the energy of ground state will be when [tex]n_{x}[/tex],[tex]n_{y}[/tex] and [tex]n_{z}[/tex] = 1

therefore energy of ground state = [tex]\frac{h^{2} }{8m}  (\frac{1 ^{2} }{2l ^{2}  }  +\frac{1^{2}  }{l^{2} }+ \frac{1^{2}  }{l ^{2} })[/tex]

=[tex]\frac{9h^{2} }{32ml^{2} }[/tex]

Otras preguntas