Electron beam, potential barrier

[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, that's 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 ?

 

[NateD]

What would happen if we now make the potential negative so the electron beam will gain the kinetic energy? If we make the potential negative, the electron beam will gain kinetic energy. This means that more of the beam will pass through the potential barrier rather than being reflected.