Problem Statement
In a Townsend discharge, the first Townsend ionization coefficient is $\alpha = 10$ cm$^{-1}$ and the distance between electrodes is $d = 1$ cm. If the number of electrons leaving the cathode per second is $n_0 = 10^6$, find the number of electrons reaching the anode.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$n = n_0 e^{\alpha d} = 10^6 \times e^{10\times1} = 10^6 \times e^{10}$$
$$n = 10^6\times22026 = 2.2\times10^{10}$$
$$\boxed{n \approx 2.2\times10^{10}\text{ electrons/s}}$$
Answer
$$\boxed{n \approx 2.2\times10^{10}\text{ electrons/s}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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