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]
