# Units in a quantum barrier problem

1. Sep 12, 2005

### ClaesF

I have a quesion regarding a quantum physics assignemnt, I wonder what units I should use when calculating the transmission coefficient of a quantum barrier problem.

I have got the following expression:

$$T = \frac{4(E+V_0)}{(2E+V_0)cos^2a\sqrt{\frac{2m}{\hbar^2}(E-V_0)} + (E-V_0+\frac{E(E+V_0)}{E-V_0}+2\sqrt{E(E+V_0)})sin^2a\sqrt{\frac{2m}{\hbar^2}(E-V_0)}}$$

where

T = the transmission coefficient
E = the energy of an incoming electron = 2.1 eV
$V_0$ = a potential = 1.5 eV
a = a point along the x-axis = 12 angstrom (= $12*10^{-10}$ m)
m = the mass of the electron (= $9.109*10^{-31}$ kg)
$\hbar$ = $1.0546*10^{-34}$ Js or $6.582*10^{-16}$ eVs.

I don't know if I should translate all values in the whole expression into SI units, or if I somehow can use the values given in the assignment in eV and angstrom directly?
If I use the eV- and angstrom values, I guess it is wrong to use the kg-value of the electronmass in the $\sqrt{\frac{2m}{\hbar^2}(E-V_0)}$-expressions.

2. Sep 12, 2005

### gnpatterson

If you use one of the planks constants in one form (Js) and one in the other form (eVs) in the h^2 expression it should work out OK.

If you think about getting the final expression inside the trig functions to be unitless. Remember J=kg m/s

3. Sep 12, 2005

### ClaesF

ok, but I must still express a in meters(=$12*10^{-10}$m) then. (Js=$kgm^2/s$)

thanks for the help.

/Claes

4. Sep 12, 2005

### Gokul43201

Staff Emeritus
You want the argument in the trig terms to be unitless. Write "a" in meters and the terms inside the square roots in SI units (so the units of the wave vector will be 1/meter).

You can leave the energies outside the trig terms in eV or SI, since the transmission coefficient is the ratio of these energy terms.