Verified Solution • 2025–26 Session moderate

QPages 395-396: Q19: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q19 Solution: The wavelengths of and X-rays of a material are 21.3 pm and 141 …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently add the given wavelengths (e.g., λ_Kβ = λ_Kα + λ_Lα) instead of adding the corresponding photon energies; the correct relation is E_Kβ = E_Kα + E_Lα, so each wavelength must first be converted to energy (or frequency) using $E=hc/\lambda$, with proper unit conversion, before summing and converting back to a wavelength.

The wavelengths of KαK_\alpha and LαL_\alpha X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of KβK_\beta X-ray of the material. Use Planck constant h=6.63×10−34 Js=4.14×10−15 eVsh = 6.63 \times 10^{-34} \text{ Js} = 4.14 \times 10^{-15} \text{ eVs}, speed of light c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.)

Step-by-Step Verified Solution

**Answer:** Wavelength of KβK_\beta X-ray = 18.5 pm. **Step-by-Step Explanation:** Given: λ(Kα)=21.3\lambda(K_\alpha) = 21.3 pm, λ(Lα)=141\lambda(L_\alpha) = 141 pm. Energies:
EKα=124021.3×10−3=58.2 keVE_{K\alpha} = \frac{1240}{21.3 \times 10^{-3}} = 58.2 \text{ keV}
ELα=1240141×10−3=8.79 keVE_{L\alpha} = \frac{1240}{141 \times 10^{-3}} = 8.79 \text{ keV}
From energy level relations:
EKβ=EKα+ELα=58.2+8.79=66.99 keVE_{K\beta} = E_{K\alpha} + E_{L\alpha} = 58.2 + 8.79 = 66.99 \text{ keV}
λ(Kβ)=124066990=18.5 pm\lambda(K_\beta) = \frac{1240}{66990} = 18.5 \text{ pm}

All Questions in this Chapter

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