Verified Solution • 2025–26 Session moderate

QPages 384-386: Q38: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q38 Solution: The light emitted in the transition to in hydrogen is called H…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently reverse the order of the quantum numbers and write ΔE = 13.6(1/n_i^2 - 1/n_f^2), which gives a negative or incorrect magnitude; remember that photon energy is the drop in energy, so use ΔE = 13.6(1/n_f^2 - 1/n_i^2) with n_i>n_f to obtain a positive value before comparing to the work function.

The light emitted in the transition n=3n = 3 to n=2n = 2 in hydrogen is called Hα_\alpha light. Find the maximum work function a metal can have so that Hα_\alpha light can emit photoelectrons from it.

Step-by-Step Verified Solution

**Answer:** Maximum work function = 1.89 eV. **Step-by-Step Explanation:** H_ light (n=3_2) has energy E = 13.6(1/4 _ 1/9) = 1.89 eV. For photoelectric emission, work function must be less than 1.89 eV [citation:3].

All Questions in this Chapter

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