An electron makes five crests during one revolution round H-nucleus. The electron belongs from the — Atomic Structure Chemistry Question

💡 Solution & Explanation

**Step 1 - De Broglie Standing Wave Condition** According to de Broglie's wave-particle duality, the circumference of a stable Bohr orbit must accommodate an integral number of de Broglie wavelengths: $$2\pi r = n\lambda$$ Where $n$ is the principal quantum number (orbit number) = number of complete waves = number of crests. --- **Step 2 - Relating Crests to Orbit Number** A complete wave has exactly one crest. So: $$\text{Number of crests} = n$$ Given: electron makes **5 crests** in one revolution: $$n = 5$$ --- **Step 3 - Identify the Orbit** Since $n = 5$, the electron belongs to the **fifth orbit**: $$\text{Orbit} = \boxed{\text{fifth orbit (n = 5)}}$$ --- **Step 4 - Evaluation of Options** * **Option (A) first orbit:** $n=1$ → only 1 crest. Incorrect. * **Option (B) fourth orbit:** $n=4$ → 4 crests. Incorrect. * **Option (C) fifth orbit:** $n=5$ → 5 crests. Correct. * **Option (D) sixth orbit:** $n=6$ → 6 crests. Incorrect.