Particle in a well, can someone explain how this step works

[Theodore0101]

Homework Statement

A particle with mass m is effected by potential V(x) = ∞ when x<0, -V0 when 0<x<a, 0 when x>a
A) Set up a relation from which the energy for bounded states can de determined
B) For which values V0 does the particle have only one bounded state?

Relevant Equations

tan(√(2ma^2(E+V0)/h^2)=-√(E+V0/-E)

Homework Statement:: A particle with mass m is effected by potential V(x) = ∞ when x<0, -V0 when 0<x<a, 0 when x>a
A) Set up a relation from which the energy for bounded states can de determined
B) For which values V0 does the particle have only one bounded state?
Homework Equations:: tan(√(2ma^2(E+V0)/h^2)=-√(E+V0/-E)

Hi! I understand the most of this example problem, I have no problem with A), which the answer is the relation tan(√(2ma^2(E+V0)/h^2)=-√(E+V0/-E), and I mostly understand B), it is just one part in the end I can't follow. Choosing the tangens function so that it only has one period in the corresponding intervall -V0<E<0 and the equations only have one point that they meet in. Therefor we put that pi/2 < √(2ma^2(E+V0)/h^2) <3pi/2 since pi/2 to 3pi/2 is a period for a tangens function, but from there they say that the equation gives pi^2*h^2/8ma^2 < V0 < 9pi^2*h^2/8ma^2, but I don't follow this step. What happened to E, why are you allowed to take it away? Shouldn't it be pi^2*h^2/8ma^2 < V0 + E < 9pi^2*h^2/8ma^2 (instead of just V0)?

Thanks

 

[Abhishek11235]

Are you sure it is $\tan$ function only? Please check your solution. For that you don't need the restriction π/2 to 3π/2. The curve will always intersect the graph between 0 to π/2. See the graph below.

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If you meant $\cot$ then the restriction applies. In that case simply note that since by defination of bound state $-V_0 \leq E \leq 0$, $E+V_0$ is positive only if E=0 for then only inequalty is valid. See below.

In both the graphs I have changed variables for my convenience.