Question

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.

Answer 1

Answer:

the energy of groud state = $\frac{9h^{2} }{32ml^{2} }$

Explanation:

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

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

therefore the energy of ground state will be when $n_{x}$,$n_{y}$ and $n_{z}$ = 1

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

=$\frac{9h^{2} }{32ml^{2} }$