# Simple atomic unit conversion check

1. Jun 3, 2015

### jjr

Hi

I'm converting some numbers to atomic units, and it's important that I get it right. Would be happy if someone could reassure me.

I want to convert the units of an electric field measured in kV/cm to atomic units. Here is what I got:

$\frac{kV}{cm}=\frac{10^3}{10^{-2}}\frac{J}{C\cdot m}=10^5\frac{2.29\cdot 10^{17}}{1.60\cdot10^{19}5.29\cdot10^{11}}a.u.=2.71\cdot 10^{-9}a.u.$

Sincerely,
J

2. Jun 3, 2015

### Dchair

No, I don't believe that is right. I did the calculation 2 different ways, and got 1 kV/cm = 1.945⋅10-7 a.u. both ways.

1st method: dimensional analysis. I believe the atomic unit for electric field is Eh/(e⋅ao), or Hartree per fundamental charge per Bohr radius.
1 kV/cm * 1000 V/kV * 100 cm/m = 105 V/m (I agree with you up to here), = 105 eV/(e*m).
105 eV/(e*m) * 1/27.2 Eh/eV * 5.29e-11 m/ao = 1.945e-7 Eh/(e*ao)

2nd method: NIST Reference (a reliable source) says that the atomic unit of electric field is 5.1422e11 V/m.
1e5 V/m * 1/5.142e11 a.u./(V/m) = 1.945e-7 a.u.

Reference: http://physics.nist.gov/cgi-bin/cuu/Value?auefld

3. Jun 4, 2015

### jjr

Yeah, it seems you are correct. Thanks!