Problem 6: Potential Difference for Neon Discharge
Given:
- Ionization energy of Neon $E_0 = 21.5 \text{ eV}$
- Plate separation $d = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$
- Mean free path $l = 0.4 \text{ mm} = 0.4 \times 10^{-3} \text{ m}$
- Charge of electron $e = 1.6 \times 10^{-19} \text{ C}$
Reasoning: For a discharge to initiate, an electron must accelerate in the electric field and gain enough kinetic energy to ionize a neon atom upon collision. This energy gain must occur within the distance of a mean free path ($l$).
The energy gained by an electron traveling distance $l$ in electric field $\mathcal{E}$ is:
$$ \text{Work} = F \cdot l = (e\mathcal{E}) \cdot l $$We require this work to equal the ionization energy $E_0$:
$$ e \left( \frac{V}{d} \right) l = E_0 $$Solving for the potential difference $V$:
$$ V = \frac{E_0 d}{e l} $$Substituting values:
$$ V = \frac{(21.5 \text{ eV}) (4 \text{ mm})}{e (0.4 \text{ mm})} $$Note: The $e$ in the denominator cancels the $e$ in the unit eV.
$$ V = \frac{21.5 \times 4}{0.4} = 215 \text{ V} $$
Answer: $$ \frac{E_0 d}{el} $$