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QPages 384-386: Q32: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q32 Solution: Electrons are emitted from an electron gun at almost zero…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often take the 656.3 nm photon energy (~1.9 eV) as the kinetic energy the electron must have, but the colliding electron must supply the full excitation energy from the ground state to n=3 (ΔE = 13.6 eV(1‑1/9) ≈ 12.1 eV). Use the Bohr formula for the level difference, not the emitted photon wavelength, to compute the required field.

Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?

Step-by-Step Verified Solution

**Answer:** Minimum electric field E = 12.1 V/m. **Step-Step Explanation:** Red light (656.3 nm) corresponds to n=3_2 transition. To emit this, the atom must be excited to n=3, requiring 12.09 eV. Electron accelerated through distance 1 m gains energy eV = 12.09 eV, so V = 12.09 V. E = V/d = 12.09/1 = 12.1 V/m [citation:3].

All Questions in this Chapter

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