Verified Solution • 2025–26 Session moderate

QPages 395-396: Q25: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q25 Solution: A free atom of iron emits X-rays of energy 6.4 keV. Calculate…

A free atom of iron emits KαK_\alpha X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of an iron atom = 9.3×10−26 kg9.3 \times 10^{-26} \text{ kg}. 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:** Recoil kinetic energy = 3.7×10−43.7 \times 10^{-4} eV. **Step-by-Step Explanation:** Given: E=6.4E = 6.4 keV, mass of iron atom M=9.3×10−26M = 9.3 \times 10^{-26} kg. Photon momentum: p=E/cp = E/c. Recoil kinetic energy of atom:
K=p22M=E22Mc2=(6.4×103×1.6×10−19)22×9.3×10−26×(3×108)2K = \frac{p^2}{2M} = \frac{E^2}{2Mc^2} = \frac{(6.4 \times 10^3 \times 1.6 \times 10^{-19})^2}{2 \times 9.3 \times 10^{-26} \times (3 \times 10^8)^2}
K=(1.024×10−15)22×9.3×10−26×9×1016=1.048×10−301.674×10−8=6.26×10−23 JK = \frac{(1.024 \times 10^{-15})^2}{2 \times 9.3 \times 10^{-26} \times 9 \times 10^{16}} = \frac{1.048 \times 10^{-30}}{1.674 \times 10^{-8}} = 6.26 \times 10^{-23} \text{ J}
K=6.26×10−231.6×10−19=3.9×10−4 eVK = \frac{6.26 \times 10^{-23}}{1.6 \times 10^{-19}} = 3.9 \times 10^{-4} \text{ eV}

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