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QPages 384-386: Q24: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q24 Solution: Average lifetime of a hydrogen atom excited to state is s. Find …
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often insert $n=2$ into an incorrect period formula that lacks the $n^3$ (or $n^4$) scaling, producing a period that is many orders of magnitude too large and thus a tiny revolution count. Use the proper Bohr orbital period $T_n=\frac{2\pi\hbar^3}{Z^2e^4m_e}\,n^3$ (or its equivalent $T_n=\frac{2\pi a_0}{\alpha c}\,n^3$) before dividing the $10^{-8}\,$s lifetime to obtain the correct number of revolutions.
Average lifetime of a hydrogen atom excited to state is s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.
Step-by-Step Verified Solution
**Answer:** Number of revolutions = 8.2.
**Step-by-Step Explanation:**
For n = 2, the time period is T = 4___n_h_/(me_) = 1.22 _ 10___ s.
Average lifetime = 10__ s.
Number of revolutions = 10__ / (1.22 _ 10___) = 8.2 _ 10_. Standard: 8.2 [citation:14].
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