Solution to equation 
	κ=ltan(la) 
for (Subscript[V, 0])=25  for even-parity solution in terms of E. Note that there are only three solutions for this choice of (Subscript[V, 0]).
κ=(Sqrt[-((2m E)/(ℏ^2))]), l =(Sqrt[((2m (E-(Subscript[V, 0])))/(ℏ^2))])

hbar=1;V0=25.0;m=1;a=1.0;
l[ee_]:=Sqrt[2m(ee+V0)]/hbar;
k[ee_]:=Sqrt[-2m ee]/hbar;
fl[ee_]:=l[ee]*Tan[a*l[ee]];
Plot[{k[ee],fl[ee]},{ee,0,-3V0},AxesLabel->{"E","y=l;y=κ"},PlotLabel->
"Soltion to  κ=ltan(la) for (Subscript[V, 0])="<>ToString[V0]]

hbar=1;V0=500.0;m=1;a=1.0;
l[ee_]:=Sqrt[2m(ee+V0)]/hbar;
k[ee_]:=Sqrt[-2m ee]/hbar;
fl[ee_]:=l[ee]*Tan[a*l[ee]];
Plot[{k[ee],fl[ee]},{ee,0,-2V0},AxesLabel->{"E","y=l;y=κ"},PlotLabel->
"Soltion to  κ=ltan(la) for (Subscript[V, 0])="<>ToString[V0]]

Solution to equation κ=ltan(la) for(Subscript[V, 0])=500  for even-parity solution.Note that there are 11 solutions for this choice of(Subscript[V, 0]).

Solution to tan z=(Sqrt[(((((Subscript[z, 0])/z)))^2)-1])for (Subscript[V, 0])=500, 50, 5 unit. Note that as (Subscript[V, 0])⟶0, there is only one solution left. It is located in the range of 0 < z < π/2. For small z, the roots tend to occur near 
to the values of nπ/2.

hbar=1;m=1;a=1.0;
z0=(a/hbar)Sqrt[2m V0];
f2[z_]:=Sqrt[(z0/z)^2-1];
Plot[{f2[z]/.{V0->500},f2[z]/.{V0->50},f2[z]/.{V0->1},Tan[z]},{z,0,10Pi},PlotStyle->{Red,Blue,Green,Black},PlotLabel->{"Red:(Subscript[V, 0])=500;Blue:(Subscript[V, 0])=50;Green:(Subscript[V, 0])=1"},AxesLabel->{"z","y=(Sqrt[(((((Subscript[z, 0])/z)))^2) - 1]); y=tan z"}]