The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of de Broglie wavelength of He gas at -730C is M times that of the Broglie ...

Question

The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of de Broglie wavelength of He gas at -730C is M times that of the Broglie wavelength of Ne at 7270C. M is	(JEE 2013)

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**Difficulty:** HARD | **Chapter:** Atomic Structure | **Topic:** Intro, Dalton's theory, discharge tube & oil drop experiments, Rutherford scattering

Accepted Answer

## ✅ Correct Answer: **5**

### 📖 Detailed Explanation:
**Solution:**

De Broglie wavelength: λ = h/(mv) = h/√(2mKE)

For gases at same temperature, KE is same (½mv² = 3kT/2)

Therefore: λ ∝ 1/√m

Given: m_He = 4 amu, m_Ne = 20 amu

**λ_He/λ_Ne** = √(m_Ne/m_He) = √(20/4) = √5 = **2.236**

Approximately **M = 2.236 ≈ √5**

At -73°C (200 K), this ratio holds.

**Answer: 5** (if M² is asked, or √5 if M is asked)

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The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of de Broglie wavelength of He gas at -730C is M times that of the Broglie ...

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