# Elektron beam strikes in potential barrier

1. Apr 20, 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 $$V_{0}$$

2. Apr 20, 2010

### nickjer

Yes, the reflection coefficient will be 0.4. So just solve for $$V_0$$ using that equation and knowing E = 10 eV.

3. Apr 23, 2010

### rayman123

I have calculted $$k_{1}= \sqrt{\frac{9.1\cdot10^{-31}\cdot16.02\cdot10^{-19}}{(6.626\cdot10^{-34})^{2}}$$=$$0.2728$$

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

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 $$k_{1}= 0.2728$$ does anyone know i am correct so far?
if yes, how would you solve this ?

4. Apr 23, 2010

### nickjer

Hmm... I wouldn't use meters as the units. Try to stick in eV and Angstroms. A couple of useful constants are:

$$m_e c^2 = 0.511\,\text{MeV}$$
$$\hbar c \approx 2000\,\text{eV}\cdot\text{\AA}$$

But I get a very different answer than your $$k_1$$. Remember that the wavelength of a particle is inversely related to the $$k$$ vector. So you are saying this electron has a wavelength on the order of meters. I think that is something you'd be able to see, which doesn't seem quite right.

5. Apr 24, 2010

### rayman123

hello you are right! I forgot that $$\hbars$$ is not the same as h!!!so here is my correct calculation
$$k_{1} = \sqrt{\frac{2mE}{\hbar^2}}=\sqrt{\frac{2\cdot9.1\cdot10^{-31}\cdot10eV}{(6.582\cdot10^{-16})^2}$$ =$$6.48$$ is this value correct

i have substituted m in [kg] and $$\hbar$$ in eVs

why do you mention about Ångström? Do i need to use this unit in any of these calculations?

6. Apr 24, 2010

### nickjer

You converted all the units to SI units except for energy. Why did you leave it as eV? Also what are the units of your $$k_1$$ in that problem.

7. Apr 24, 2010

### rayman123

the unit for k should be $$\frac{1}{m}$$can i express energy in J instead and do the same thing with $$\hbar$$ and express it in Js?

8. Apr 24, 2010

### nickjer

If you want to solve it in SI units then you need to keep everything in SI units. So yes, convert eV to J, and make sure $$\hbar$$ is in SI units.

9. Apr 24, 2010

### rayman123

in that case i get very big number.....
$$\sqrt{\frac{2\cdot9.1\cdot10^{-31}\cdot1.602\codot10^{-18}}{(1.054\cdot10^{-34})^2}$$= $$\sqrt{\frac{2.91564\cdot10^{-48}}{1.110916\cdot10^{-68}}$$ =$$\sqrt{2.624\cdot10^{20}}$$ = $$1.62\cdot10^{10}$$

which is extraordinary big.....dont you think?

10. Apr 24, 2010

### nickjer

It seems very reasonable. Remember that the wavelength of the electron is inversely proportional to the k-vector. So that looks like a wavelength on the order of an angstrom. So that is better looking than meters.

11. Apr 24, 2010

### rayman123

okey i am glad to hear that.
do you have any idea how to solve the equation for R to obtain $$k_{2}$$ (which should be the next step to calculate $$V_{0}$$)?

$$R= \frac{(k_{1}-k_{2})^2}{(k_{1}+k_{2)})^2}$$

$$R(k_{1}+k_{2})^2=(k_{1}-k_{2})^2$$

$$(\sqrt{R}k_{1}+\sqrt{R}k_{2})^2-(k_{1}-k_{2})^2=0$$

$$\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}-k_1+k_2\right)\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}+k_1-k_2\right)=0$$

$$\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}-k_1+k_2=0\right)\quad\vee\quad\left(\sqrt{R}k_{1}+\sqrt{R}k_{2}+k_1-k_2=0\right)$$

$$\left(-k_1(1-\sqrt{R})+k_2(1+\sqrt{R})=0\right)\quad\vee\quad\left(k_1(1+\sqrt{R})-k_2(1-\sqrt{R})=0\right)$$

$$\left(k_2(1+\sqrt{R})=k_1(1-\sqrt{R})\right)\quad\vee\quad\left(k_1(1+\sqrt{R})=k_2(1-\sqrt{R})\right)$$

$$\left(k_2=k_1\frac{(1-\sqrt{R})}{(1+\sqrt{R})}\right)\quad\vee\quad\left(k_1\frac{(1+\sqrt{R})}{(1-\sqrt{R})}=k_2\right)$$

but then i get 2 values for $$k_{2}$$

Last edited: Apr 25, 2010