# De Broglie wave length of electron [closed]

Consider an electron with total energy $E>V_2$ in a potential well with $$V(x)= \begin{cases} \infty & x< 0 \\ V_1 & 0< x< L \\ V_2 & x>L \end{cases}$$ where $V_2>V_1>0$.

We can determine that $$\phi_E(x)= \begin{cases} 0 & x< 0 \\ A\sin(kx)+B\cos(kx) & 0< x< L \\ Ce^{qx}+De^{-qx} & x>L \end{cases}$$ where $k^2=\frac{2m_e E}{\hbar}$ and $q=k\sqrt{V_1-E}$.

We can also apply the boundary condition at $x=0$ to determine that $$\phi_e(x)=A\sin(kx)$$ for $x\in[0,L]$.

We can also apply boundary conditions at $x=L$ to find that $$A\sin(kL)=De^{-qL}$$ $$Ak\cos(kL)=-Dqe^{-qL}$$ (since $C=0$ due to the corresponding positive exponent), and $$k\cot(kL)=-q$$

I need to calculate the de Broglie wavelength of the electron in the regions defined by the potential, if $V_1 = 10.0$ eV, $V_2 = 20.0$ eV, $E=30$ eV.

For the region where $0\le x\le L$, if I'm correct, the de Broglie $\lambda$ is determined by $\frac{h}{p}$, where $p=\hbar k=\sqrt{2m_e E}$, and so we can find $\lambda_{\text{de Broglie}}=8.962639\times 10^{-18}$ m.

For the region where $x<0$, we can't find this length because of the infinite potential.

For the region where $x>L$, if $p=\hbar q$, then it appears that $p$ is complex-valued, and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.

Would appreciate some help.

## closed as off-topic by CuriousOne, user36790, ACuriousMind♦, honeste_vivere, John RennieJul 20 '16 at 16:45

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• It's not actually very clear to me what the question here is - you say you would "appreciate some help" but what is your question? Just how to calculate the deBroglie wavelength correctly? – ACuriousMind Jul 20 '16 at 13:37

## 1 Answer

As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts represent the incoming and reflected waves respectively. That's essentially the wave function of a free particle.

Its de Broglie wave length where $x>L$, is calculated using:

$$p=\hbar k=\sqrt{2m_e (E-V_2)}$$

But for bound particles with $E<V_2$, the wave function $\psi$ (noted as $\phi$ by you) is Real over the entire domain $0\leq x \leq+\infty$.

For $0 \leq x \leq L$, then $p=\hbar k=\sqrt{2m_e (E-V_1)}$ (bound particle).

and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.

You're entirely correct to drop $C$ ($C=0$) for bound particles but note that not doing so wouldn't make $\psi(x)$ Complex in that area of the domain. Classically a particle with $E<V_2$ could not enter into that $x \leq L$ area but a quantum particle tunnels into the classically forbidden area.

1D SE solution for free particle of energy $E$ in constant potential field $V$: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ $$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V)\psi=0$$ $$k^2=\frac{2m}{\hbar^2}(E-V)$$ $$\frac{d^2\psi}{dx^2}+k^2\psi=0$$ For: $$k>0$$ $$\implies \psi(x)=c_1e^{-ikx}+c_2e^{+ikx}$$

• Since the particle is unbound, did I need to drop the coefficient $C$? I think the function should decay in $x>L$. Also, for $x>L$, $q=k\sqrt{V_1-E}$, not $k\sqrt{E-V_1}$ do you think this is correct? – sequence Jul 20 '16 at 3:19
• For the unbound particle: 1. $0\leq x \leq$, then $k=\sqrt{2m_e (E-V_1)}/\hbar$. 2. $x>L$, then $k=\sqrt{2m_e (E-V_2)}/\hbar$. The amplitude of the wave function drops a bit when going from $V_1$ to $V_2$. However, it does not decay further. Solve the SE for a free particle in a potential $V$ (with $V<E$). $C \neq 0$. – Gert Jul 20 '16 at 12:13
• Ooopsie. That should have read the wave number drops a bit when [...]. Not the amplitude. – Gert Jul 20 '16 at 15:26