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QPages 384-386: Q16: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q16 Solution: A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently take the calculated photon energies ($E=hc/\lambda$) as the absolute energies of the atomic states, forgetting that each wavelength represents a *difference* between two levels. Set up $E_3-E_2$, $E_3-E_1$, and $E_2-E_1$ using the three observed wavelengths (with $E_3=0$) and solve for $E_2$ and $E_1$, respecting the decreasing spacing rule, to avoid this error.

A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

Step-by-Step Verified Solution

**Answer:** Ground state energy = _27 eV, first excited state energy = _12 eV. **Step-by-Step Explanation:** The wavelengths are 46.0 nm, 82.8 nm, and 103.5 nm. Energies: E_ = 1240/46.0 = 26.96 eV, E_ = 1240/82.8 = 14.98 eV, E_ = 1240/103.5 = 11.98 eV. The highest energy state is taken as zero. The three energies correspond to transitions between three levels. The ground state is the lowest, so its energy relative to the highest state is _27 eV (approximately). The first excited state is _12 eV. Standard answer: ground state = _27 eV, first excited state = _12 eV [citation:6].

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