Kvant Physics Problem 40
A neon-filled discharge tube is placed between two large parallel plate electrodes separated by a distance
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m52s
Source on kvant.digital
Problem
At what potential difference between the electrodes does a neon lamp ignite if the ionization energy of neon is $I=21{,}5~\text{эВ}$, and the average 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 separated by a distance of $d=3~\text{мм}$ from each other.
Setup and Assumptions
A neon-filled discharge tube is placed between two large parallel plate electrodes separated by a distance
$$d=3,\text{mm}=3.0\times10^{-3},\text{m}.$$
The average distance traveled by an electron between successive collisions with neon atoms is
$$\lambda=0.4,\text{mm}=4.0\times10^{-4},\text{m}.$$
The ionization energy of a neon atom is
$$I=21.5,\text{eV}.$$
The unknown quantity is the potential difference $U$ between the electrodes at which the gas discharge begins.
The electric field between the plates is assumed uniform because the plates are large compared with their separation. The electron motion is treated classically between collisions. Energy losses other than those occurring in collisions are neglected. The gas is assumed sufficiently rarefied that the mean free path remains equal to $\lambda$. Ignition is assumed to occur when an electron can acquire at least the ionization energy of a neon atom during one free flight between successive collisions.
Physical Principles
In a uniform electric field $E$, the work done by the field on a charge $e$ over a distance $l$ is
$$A=eEl.$$
The potential difference and electric field between parallel plates are related by
$$E=\frac{U}{d}.$$
For ionization to occur during a free flight, the kinetic energy gained by an electron between collisions must be at least equal to the ionization energy:
$$eE\lambda = I.$$
Since the ionization energy is given in electron-volts, the quantity $eE\lambda$ expressed in electron-volts is numerically equal to the potential drop over the mean free path:
$$E\lambda = I \quad (\text{in volts}).$$
Derivation
The condition for the onset of ionization is
$$eE\lambda = I.$$
Substituting
$$E=\frac{U}{d},$$
gives
$$e\frac{U}{d}\lambda = I.$$
Solving for the required voltage,
$$U=\frac{Id}{e\lambda}.$$
Because $I$ is specified in electron-volts, division by $e$ converts the energy directly into volts:
$$\frac{I}{e}=21.5,\text{V}.$$
Hence
$$U=21.5,\text{V}\cdot\frac{d}{\lambda}.$$
Substituting the given distances,
$$U=21.5,\text{V}\cdot \frac{3.0,\text{mm}}{0.4,\text{mm}}.$$
The millimeters cancel:
$$U=21.5\times7.5,\text{V}.$$
Therefore,
$$U=161.25,\text{V}.$$
Result
The ignition voltage is
$$U=\frac{Id}{e\lambda}.$$
Numerically,
$$U=21.5,\text{V}\cdot\frac{3.0,\text{mm}}{0.4,\text{mm}} =161.25,\text{V}.$$
Thus,
$$\boxed{U \approx 1.6\times10^{2}\ \text{V}}$$
or, to sensible significant figures,
$$\boxed{U \approx 160\ \text{V}}.$$
Sanity Checks
The ratio $d/\lambda=3.0/0.4=7.5$ is dimensionless, so the final expression has units of volts, as required.
If the mean free path were larger, an electron would gain more energy between collisions, and the required voltage would decrease. The formula
$$U\propto \frac{1}{\lambda}$$
has exactly this behavior.
If the electrode spacing were increased while keeping $\lambda$ fixed, the electric field produced by a given voltage would decrease, so a larger voltage would be required. The formula
$$U\propto d$$
matches this expectation.
The most common source of error is mixing electron-volts and joules. Using $I=21.5,\text{eV}$ directly as if it were $21.5,\text{J}$ would produce a completely incorrect result. Another common mistake is using the full electrode spacing $d$ instead of the mean free path $\lambda$ in the ionization condition. The electron must acquire the ionization energy during a single free flight, so the relevant distance is $\lambda$, not $d$.