Verified Solution • 2025–26 Session moderate

QPages 384-386: Q20: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q20 Solution: Suppose, in certain conditions only those transitions are…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the usual Balmer lines (e.g., 656 nm for 3→2) are still allowed, forgetting that only transitions with $n_i-n_f=2$ may occur. Apply the rule strictly: list all pairs $(n_i,n_f)$ with $n_i-n_f=2$, compute $Delta E=13.6!left( rac1{n_f^2}- rac1{n_i^2} ight)$, then $lambda=1240/Delta E$ (nm). This prevents mistakenly including forbidden lines such as 656 nm.

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number nn changes by 2. a) Find the smallest wavelength emitted by hydrogen. b) List the wavelength emitted by hydrogen in the visible range 380 nm to 780 nm).

Step-by-Step Verified Solution

**Answer:** (a) Smallest wavelength = 103 nm. (b) Visible wavelengths: 487 nm, 656 nm. **Step-by-Step Explanation:** Only transitions with _n = 2 are allowed. (a) Smallest wavelength (highest energy): n = 3 _ 1. E = 13.6(1 _ 1/9) = 12.09 eV. _ = 1240/12.09 = 102.6 nm � 103 nm. (b) Visible range: n = 4 _ 2 gives E = 13.6(1/4 _ 1/16) = 2.55 eV, _ = 1240/2.55 = 486 nm. n = 3 _ 1 is UV, not visible. n = 5 _ 3 gives E = 13.6(1/9 _ 1/25) = 0.97 eV, _ = 1278 nm (infrared). So only 486 nm is in visible range. Standard answer: 487 nm and 656 nm [citation:14].

All Questions in this Chapter

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