(a) Here the schrodinger equation is
`(ħ^(2))/(2m)((del^(2))/(delx^(2))+(del^(2))/(dely^(2)))Psi=E Psi`
We take the origin at one of the corners of the rectangle where the particles can lie. Then the wave function must vanish for
`x=0 or x= l_(1)`
or `y=0 or y=l_(2)`
We look for a solution in the form
`Psi= A sin k_(1) x sin k_(2)y`
cosines are not permitted by the boundary conditions. Then
`k_(1)=(n_(i)pi)/(l_(1)),k_(2)=n_(2)(pi)/(l_(2))`
and `E=(k_(1)^(2)+k_(2)^(2))/(2m)ħ^(2)=(pi^(2)ħ^(2))/(2m)((n_(1)^(2))/(t_(1)^(2))+(n_(2)^(2))/(t_(2)^(2)))`
Here, `n_(1),n_(2)` are nonzero integers
(b) If `l_(1)=l_(2)=1` then
`(E)/(ħ^(2)//ml^(2))=(n_(1)^(2)+n_(2)^(2))/(2)pi^(2)`
1st level: `n_(1)=n_(2)=2= 1rarrpi^(2)=9.87`
`2^(nd)` level: `,{:(n_(1),=,n_(2),=,2),(or n_(1),=2,n_(2),=,1):}}rarr(5)/(2)pi^(2)= 24.7`
`3^(st)" level": n_(1)=2,n_(2)=2 rarr 4pi^(2)= 39.5`
`4^(nd)" level":{:(n_(1)1,=,n_(2),=,3),(n_(1),=3,n_(2),=,1):}}rarr 5pi^(2)= 49.3`