I understand the relationships regarding strength of the electric field and its

Question

I understand the relationships regarding strength of the electric field and its pressure/temperature, I am just having trouble calculating the actual value for the atmospheric electric field. Here is the stated question:

Sparks in air occur when ions in the air are accelerated to such a high speed by an electric field that when the ions impact on neutral gas molecules, the neutral molecules become ions. If the electric field strength is large enough, the ionized collision products are themselves accelerated and produce more ions on impact, and so forth. This avalanche of ions is what we call a spark.

(a) Assume that an ion moves, on average, exactly one mean free path through the air before hitting a molecule. If the ion needs to acquire approximately 1.13 eV of kinetic energy in order to ionize a molecule, estimate the minimum strength of the electric field required at standard room pressure and temperature (300 K). Assume that the cross-sectional area of an air molecule is about 0.101 nm2. (Assume the velocities of the particles have a Maxwell distribution of velocities.)
-----

The answer I have obtained multiple times was 2.79x10^6 N/C, but apparently this is incorrect. Any help here would be appreciated.

Explanation / Answer

KE = Force * distance Force = q * E E = electric-field strength = 3e6 (N/C) q = electron's charge = 1.6*10^(-19) (C) Therefore: distance = KE/Force = 3e(-19)/(q*3e6) = 3e(-19)/(3e6 * 1.6e(-19)) = 3/(3*1.6) * e(-6) = 6.25e(-7) Distance = 6.25*10^(-7) (m)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---