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QPages 352-354: Q1: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q1 Solution: A discharge tube contains helium at low pressure. A large…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often replace the helium ion’s mass with “4 $m_e$” because the atom has mass number 4, forgetting that $m_e$ is the electron mass (≈1/1836 of a proton). The ion’s mass is actually $\approx4\,\text{u}\approx4\times1836\,m_e$, so its acceleration is $a_{ion}=eE/m_{ion}$ and the distance ratio $s_e/s_{ion}=m_{ion}/m_e\gg4$. Use the correct ion mass to avoid the 4:1 error.

A discharge tube contains helium at low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionised due to the detachment of an atomic electron. Find the ratio of the distance travelled by the free electron to that by the positive ion in a short time dt after ionisation.

Step-by-Step Verified Solution

**Answer:** The ratio of distance travelled by the electron to that by the positive ion is 4 : 1. **Step-by-Step Explanation:** For the electron: acceleration ae=eEmea_e = \frac{eE}{m_e}. Distance in time dtdt: se=12aedt2s_e = \frac{1}{2} a_e dt^2. For the positive ion (He_): mass mp≈4mem_p \approx 4 m_e (since helium ion has mass number 4). Acceleration ap=eE4mea_p = \frac{eE}{4m_e}. sp=12apdt2=12⋅eE4medt2=se4s_p = \frac{1}{2} a_p dt^2 = \frac{1}{2} \cdot \frac{eE}{4m_e} dt^2 = \frac{s_e}{4}. Hence, se:sp=4:1s_e : s_p = 4 : 1 .

All Questions in this Chapter

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