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blaksheep423

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I have an exam later today, and I would really apreciate it if someone could check my work or at least point out flaws in my process here. Thanks!

At x=0, a proton with a kinetic energy of 10 eV is traveling in the x direction (potential energy = 0). At x=1nm, it encounters a potential barrier of height 12eV and width 0.2 nm. The potential returns to 0 at 1.2nm.

Calculate the transmission and reflection probabilities

T = [1 + [V

R = 1 - T

where V = 12, E = 10, L = 2*10

and k = [tex]\sqrt{}2m (V-E)[/tex]/h

well its all straightforward, and I got k = 7.24 * 10

which means kL is 1.45.

when I plug everything in, I get T = .12 and R = .88

I've checked it 3 times and it comes out the same every time, but this seems like a very high transmission probability, so I don't know if i made a mistake somewhere.

An electron is in a 1D potential well approximated by V = 0 for 0nm < x < 2nm and V = infinity for all other x

what is the electron wave function for this lowest energy state?

what is the energy of the electron in this state

shroedinger wave equation was used to derive this wave function

well I got [tex]\Psi[/tex] = Asin(kx), where kL = n[tex]\pi[/tex], so

[tex]\Psi[/tex] = Asin(n[tex]\pi[/tex]x / L)

when i normalize this, i get A = [tex]\sqrt{}2/L[/tex], so

[tex]\Psi[/tex] = [tex]\sqrt{}2/L[/tex]sin(n[tex]\pi[/tex]x / L)

so I plug in L and i end up with a numerical equation of

[tex]\Psi[/tex] = 31623sin (5[tex]\pi[/tex] * 10

this doesn't seem right to me.

for the energy i just plug the values into the equation:

E = h

and i get .094 eV, which also seems wrong.

## Homework Statement

At x=0, a proton with a kinetic energy of 10 eV is traveling in the x direction (potential energy = 0). At x=1nm, it encounters a potential barrier of height 12eV and width 0.2 nm. The potential returns to 0 at 1.2nm.

Calculate the transmission and reflection probabilities

## Homework Equations

T = [1 + [V

^{2}sinh^{2}(kL)] / [4E (V - E)]R = 1 - T

where V = 12, E = 10, L = 2*10

^{-10},and k = [tex]\sqrt{}2m (V-E)[/tex]/h

_{bar}## The Attempt at a Solution

well its all straightforward, and I got k = 7.24 * 10

^{9},which means kL is 1.45.

when I plug everything in, I get T = .12 and R = .88

I've checked it 3 times and it comes out the same every time, but this seems like a very high transmission probability, so I don't know if i made a mistake somewhere.

## Homework Statement

An electron is in a 1D potential well approximated by V = 0 for 0nm < x < 2nm and V = infinity for all other x

what is the electron wave function for this lowest energy state?

what is the energy of the electron in this state

## Homework Equations

shroedinger wave equation was used to derive this wave function

## The Attempt at a Solution

well I got [tex]\Psi[/tex] = Asin(kx), where kL = n[tex]\pi[/tex], so

[tex]\Psi[/tex] = Asin(n[tex]\pi[/tex]x / L)

when i normalize this, i get A = [tex]\sqrt{}2/L[/tex], so

[tex]\Psi[/tex] = [tex]\sqrt{}2/L[/tex]sin(n[tex]\pi[/tex]x / L)

so I plug in L and i end up with a numerical equation of

[tex]\Psi[/tex] = 31623sin (5[tex]\pi[/tex] * 10

^{8}x)this doesn't seem right to me.

for the energy i just plug the values into the equation:

E = h

^{2}n^{2}[tex]\pi[/tex]^{2}/ 2mL^{2}and i get .094 eV, which also seems wrong.

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