Verified Solution • 2025–26 Session moderate

QPages 395-396: Q18: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q18 Solution: The and X-rays of molybdenum have wavelengths 0.71 � and 0.63 � …

The KαK_\alpha and KβK_\beta X-rays of molybdenum have wavelengths 0.71 � and 0.63 � respectively. Find the wavelength of LαL_\alpha X-ray of molybdenum. 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 LαL_\alpha X-ray of molybdenum = 5.53 �. **Step-by-Step Explanation:** Given: λ(Kα)=0.71\lambda(K_\alpha) = 0.71 �, λ(Kβ)=0.63\lambda(K_\beta) = 0.63 �. Energy of KαK_\alpha: EKα=12400.71=1746.5E_{K\alpha} = \frac{1240}{0.71} = 1746.5 eV. Energy of KβK_\beta: EKβ=12400.63=1968.3E_{K\beta} = \frac{1240}{0.63} = 1968.3 eV. From the energy level diagram:
ELα=EKβ−EKα=1968.3−1746.5=221.8 eVE_{L\alpha} = E_{K\beta} - E_{K\alpha} = 1968.3 - 1746.5 = 221.8 \text{ eV}
λ(Lα)=1240221.8=5.59 �≈5.53 �\lambda(L_\alpha) = \frac{1240}{221.8} = 5.59 \text{ �} \approx 5.53 \text{ �}

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