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QPages 384-386: Q28: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q28 Solution: Radiation coming from transition to of hydrogen atoms falls on…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently write the hydrogen‑like level energy as $E_n=-Z^2\,13.6\,\text{eV}/n$ instead of the correct $E_n=-Z^2\,13.6\,\text{eV}/n^2$, causing non‑integer final states; always use $E_n=-13.6Z^2/n^2$ and require the photon energy to match the exact difference between two integer levels.

Radiation coming from transition n=2n = 2 to n=1n = 1 of hydrogen atoms falls on helium ions in n=1n = 1 and n=2n = 2 states. What are the possible transitions of helium ions as they absorbs energy from the radiation?

Step-by-Step Verified Solution

**Answer:** Possible transitions: He_ from n=1 to n=3 or n=4; from n=2 to n=4 or n=5. **Step-by-Step Explanation:** Radiation from hydrogen n=2_1 has energy 10.2 eV. For He_ (Z=2), energy levels are E_n = _54.4/n_ eV. From n=1 (_54.4 eV), adding 10.2 eV gives _44.2 eV, which corresponds to n_ = 54.4/44.2 = 1.23, not an integer. So no transition from n=1. From n=2 (_13.6 eV), adding 10.2 eV gives _3.4 eV, which corresponds to n_ = 54.4/3.4 = 16, so n=4. So He_ can go from n=2 to n=4 [citation:3].

All Questions in this Chapter

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