Kvant Physics Problem 40

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Problem

At what potential difference between the electrodes does a neon lamp light up, if the ionization energy of neon is $I=21{,}5~\text{эВ}$, and the mean distance between two successive collisions of an electron with gas atoms is $0{,}4~\text{мм}$? The electrodes are in the form of large plates located at a distance $d=3~\text{мм}$ from each other.

Setup and Assumptions

A neon lamp consists of two large parallel plate electrodes separated by a distance $d = 3 \times 10^{-3},\text{m}$. A uniform electric field $E$ is assumed between the plates, so the potential difference is $U = Ed$. An electron moves through neon gas and undergoes collisions with mean free path $\lambda = 0.4 \times 10^{-3},\text{m}$. Between collisions the electron is accelerated by the electric field and gains kinetic energy from the field.

The lamp ignites when an electron acquires enough energy between two successive collisions to ionize a neon atom. The ionization energy is $I = 21.5,\text{eV}$. It is assumed that ionization occurs when the energy gained over one mean free path equals $I$, and that energy losses other than collisions are neglected. The electron motion is treated in the classical regime for energy gain between collisions.

The unknown is the breakdown potential difference $U$ between the electrodes.

Physical Principles

The work done by the electric field on an electron over a distance $\ell$ is $W = eE\ell$, where $e$ is the elementary charge.

The energy gained by the electron between collisions is set by the mean free path, so the relevant distance is $\lambda$, giving $W = eE\lambda$.

Ionization occurs when this gained energy equals the ionization energy of neon, written in joules as $I e$, since $I$ is given in electronvolts.

The relation between electric field and potential difference in a uniform field is $U = Ed$.

Derivation

The energy gained by an electron between two successive collisions is

$$W = eE\lambda.$$

Ionization begins when this energy equals the ionization threshold,

$$eE\lambda = Ie.$$

Dividing both sides by $e$ gives the threshold condition in electronvolt form,

$$E\lambda = I.$$

Solving for the electric field,

$$E = \frac{I}{\lambda}.$$

The potential difference between the electrodes is related to the electric field by

$$U = Ed.$$

Substituting the expression for $E$,

$$U = \frac{I}{\lambda} d.$$

Result

Substituting the numerical values, first convert the distances:

$$\lambda = 0.4 \times 10^{-3},\text{m}, \quad d = 3 \times 10^{-3},\text{m}.$$

Compute the electric field threshold:

$$E = \frac{21.5,\text{V}}{0.4 \times 10^{-3},\text{m}} = \frac{21.5}{4 \times 10^{-4}},\text{V/m} = 5.375 \times 10^{4},\text{V/m}.$$

Then compute the potential difference:

$$U = (5.375 \times 10^{4},\text{V/m})(3 \times 10^{-3},\text{m}) = 1.6125 \times 10^{2},\text{V}.$$

$$\boxed{U \approx 1.6 \times 10^{2},\text{V}}$$

Sanity Checks

The expression $U = I d / \lambda$ has dimensions of volts because $I$ carries units of volts when expressed in electronvolts per charge, and the ratio $d/\lambda$ is dimensionless.

The result scales linearly with the plate separation $d$, so increasing electrode distance increases the required breakdown voltage, consistent with the need for a stronger total energy gain across the gap.

The field magnitude $E \sim 5 \times 10^{4},\text{V/m}$ is typical for gas breakdown conditions over millimeter-scale gaps, supporting the physical plausibility of the estimate.

The most sensitive step is equating $E\lambda = I$, since it assumes that one ionizing collision per mean free path is sufficient and neglects energy losses in non-ionizing collisions; an underestimate of losses would increase the required voltage.