# Electron beam, potential barrier

1. Apr 24, 2010

### rayman123

Electron beam with kinetic energy $$E_{k} = 10 eV$$ strikes a positive potential barrier $$V_{0}$$ and the kinetic energy after the beam has passed through the barrier is $$E_{k} = (10 eV -V_{0})$$.
How big potential $$V_{0}$$ is needed so that 40% of the electron beam is going to be reflected?
What would happen if we now make the potential negative so the electron beam will gain the kinetic energy?

I would say that the energy of the particles is higher than the energy of the potential barrier, thats why we observe transmission and reflection
solving The Schrödinger equation

$$\frac{\partial^2}{\partial x^2}\psi(x)+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0$$
then the solutions will be

$$\psi_{1}=Ae^{ik_{1}x}+Be^{-ik_{1}x}$$ in the region x<0 $$k_{1}= \sqrt{\frac{2mE}{\hbar^2}}$$
$$\psi_{1}=Ce^{ik{2}x}$$ in the region x>0 $$k_{2}= \sqrt{\frac{2m[E-V_{0}]}{\hbar^2}}$$

the reflection coefficient is $$R= (\frac{k_{1}-k_{2}}{k_{1}+k_{2}})^2$$

Can someone help me with the solution?
Is the reflection coefficient going to be 0.4? How to find that value of $$V_{0}$$[/QUOTE]

I have calculted $$k_{1}= \sqrt{\frac{9.1\cdot10^{-31}\cdot10}{(6.582\cdot10^{-16})^{2}}$$=$$6.48$$

$$k_{1}= \sqrt{\frac{2mE}{\hbar^2}}= \sqrt{\frac{kg\cdot J}{J^2\cdot s^2}$$$$=\sqrt{\frac{kg}{N^2\cdot s^2}$$=$$\frac{1}{m}$$

then i am trying to solve the equation with R

$$R= (\frac{k_{1}-k_{2}}{k_{1}+k_{2}})^2$$ to calculate $$k_{2}$$ and finally calculate $$V_{0}$$ from the equation for $$k_{2}$$

but i am stuck here....

$$R= \frac{k_{1}^2-2k_{1}k_{2}+k_{2}^2}{k_{1}^2+2k_{1}k_{2}+k_{2}^2}$$
and then i substitute for R= 04 and for $$k_{1}= 6.48$$ does anyone know i am correct so far?
if yes, how would you solve this ?
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Last edited: Apr 24, 2010