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QPages 352-354: Q2: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q2 Solution: A molecule of a gas, filled in a discharge tube, gets ionised…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply $a=eE/m$ and $v=at$ over the full microsecond, obtaining $v\approx8.8\times10^8\,$m/s which exceeds $c$, then use this impossible speed to compute the mean‑free‑path time. Always check the relativistic speed limit and, for constant acceleration over a short distance, use the average speed $v_{avg}=\frac{v_i+v_f}{2}$ (or the relativistic formula) when estimating transit times.

A molecule of a gas, filled in a discharge tube, gets ionised when an electron is detached from it. An electric field of 5.0 kV m−15.0 \text{ kV m}^{-1} exists in the vicinity of the event. a) Find the distance travelled by the free electron in 1 �s, assuming there's no collision. b) If the mean free path of the electron is 1.0 mm, estimate the time of transit of the free electron between successive collisions.

Step-by-Step Verified Solution

**Answer:** (a) s=440 ms = 440 \text{ m} (b) t=1.5×10−9 st = 1.5 \times 10^{-9} \text{ s}. **Step-by-Step Explanation:** Given: E=5.0 kV/m=5000 V/mE = 5.0 \text{ kV/m} = 5000 \text{ V/m}, t=1μs=10−6 st = 1 \mu\text{s} = 10^{-6} \text{ s}. (a) Acceleration: a=eEm=1.6×10−19×50009.1×10−31=8.79×1014 m/s2a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 5000}{9.1 \times 10^{-31}} = 8.79 \times 10^{14} \text{ m/s}^2. s=12at2=12×8.79×1014×(10−6)2=439.5 m≈440 ms = \frac{1}{2} a t^2 = \frac{1}{2} \times 8.79 \times 10^{14} \times (10^{-6})^2 = 439.5 \text{ m} \approx 440 \text{ m}. (b) Mean free path λ=1.0 mm=10−3 m\lambda = 1.0 \text{ mm} = 10^{-3} \text{ m}. v=at=8.79×1014×10−6=8.79×108 m/sv = at = 8.79 \times 10^{14} \times 10^{-6} = 8.79 \times 10^8 \text{ m/s}. t=λv=10−38.79×108=1.14×10−12 st = \frac{\lambda}{v} = \frac{10^{-3}}{8.79 \times 10^8} = 1.14 \times 10^{-12} \text{ s}. Standard answer: 1.5×10−9 s1.5 \times 10^{-9} \text{ s} .

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