Finite potential well problem when $|E| \ll V_{0}$ [hold] [closed]

The potential well looks like the following.

The problem requires the proof of the following formula in the case that $|E| \ll V_{0}$. E is the lowest energy eigenvalue.

$$\frac{\sqrt{MV_{0}}a}{\hbar} \approx \frac{\pi}{2} - \sqrt{\frac{|E|}{V_{0}}}$$

where $M$ is the mass of the particle, $a$ is the length of the box.

This is my try. [Edited according to the post by @ZeroTheHero]

In the region I, we may write the wave function with a sine function, because a cosine function doesn't satisfy the boundary condition at $x=0$.

$$\psi (x) = B \sin k x$$

where $k = \frac{1}{\hbar}\sqrt{2m (E+V_{0})}$.

In the region II, we may express the wavefunction as $$\psi(x) = Ce^{- \kappa x}$$ where $\kappa = \frac{1}{\hbar} \sqrt{-2mE}$. Only decaying exponential should be considered because the whole system is bounded.

From the boundary condition, $$B \sin ka = C e^{-\kappa a}$$ $$kB \cos ka = - \kappa C e^{- \kappa a}$$

Therefore, we get $k \cot ka = \kappa$, or

$$\cot ka = - \frac{\kappa}{k} = - \sqrt{\frac{-E}{E+V_{0}}} \approx - \sqrt{\frac{|E|}{V_{0}}}$$

Also, we have a relation, $$k^{2}+\kappa^{2} = \frac{2m V_{0}}{\hbar^{2}}$$

It seems the answer is related to some trigonometry. Any suggestion?

closed as off-topic by Kyle Kanos, heather, John Rennie, Emilio Pisanty, Jon CusterSep 19 '17 at 13:16

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The problem is with your guess solution. The infinite barrier at $x=0$ forces $\psi(x)$ to be $0$ there so you need to start from $$\psi(x)=A\sin(kx)\, ,$$ as none of the functions in $\cos(kx)$ can satisfy the boundary condition. With this adjustment things should go much more smoothly.
Edit: Write, inside the well \begin{align} \frac{\hbar^2}{2m}(k^2)&=V_0-\epsilon>0\, ,\\ k&=\sqrt{\frac{2m}{\hbar^2}(V_0-\epsilon)}\, ,\\ ka&=\sqrt{\frac{2m}{\hbar^2}V_0 a^2(1-x)}\, ,\qquad x=\frac{\epsilon}{V_0} \end{align} with $\epsilon>0$. Now, for $x\ll 1$, $$ka\approx \sqrt{\frac{2m}{\hbar^2}V_0}a\left(1-\frac{1}{2}x\right)$$ and you can work out $\cot(ka)$ by using trig identities and Taylor series fo $\cos(ka)$ and $\sin(ka)$.