Elektron beam strikes in 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}

 

[nickjer]

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

 

[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 ?

 

[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.

 

[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?

 

[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.

 

[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?

 

[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.

 

[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?

 

[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.

 

[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}