Verified Solution • 2025–26 Session moderate

QPages 395-396: Q20: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q20 Solution: The energy of a silver atom with a vacancy in K shell is 25.31…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mix the two forms of Planck's constant and forget to convert the transition energy from keV to eV (or to joules) before dividing by $h$. Use either $h=6.63\times10^{-34}\,$J·s with $E$ in joules, or $h=4.14\times10^{-15}\,$eV·s with $E$ in eV; inconsistent unit conversion yields frequencies off by a factor of $10^3$.

The energy of a silver atom with a vacancy in K shell is 25.31 keV, in L shell is 3.56 keV and in M shell is 0.530 keV higher than the energy of the atom with no vacancy. Find the frequency of KαK_\alpha, KβK_\beta and LαL_\alpha X-rays of silver. 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:** KαK_\alpha: 2.5×10182.5 \times 10^{18} Hz; KβK_\beta: 2.9×10182.9 \times 10^{18} Hz; LαL_\alpha: 7.3×10177.3 \times 10^{17} Hz. **Step-by-Step Explanation:** Given: EK=25.31E_K = 25.31 keV, EL=3.56E_L = 3.56 keV, EM=0.530E_M = 0.530 keV. Energies of transitions:
E(Kα)=EK−EL=25.31−3.56=21.75 keVE(K_\alpha) = E_K - E_L = 25.31 - 3.56 = 21.75 \text{ keV}
E(Kβ)=EK−EM=25.31−0.530=24.78 keVE(K_\beta) = E_K - E_M = 25.31 - 0.530 = 24.78 \text{ keV}
E(Lα)=EL−EM=3.56−0.530=3.03 keVE(L_\alpha) = E_L - E_M = 3.56 - 0.530 = 3.03 \text{ keV}
Frequencies:
ν=Eh=E×1.6×10−166.63×10−34\nu = \frac{E}{h} = \frac{E \times 1.6 \times 10^{-16}}{6.63 \times 10^{-34}}
KαK_\alpha: ν=5.25×1018\nu = 5.25 \times 10^{18} Hz KβK_\beta: ν=5.98×1018\nu = 5.98 \times 10^{18} Hz LαL_\alpha: ν=7.31×1017\nu = 7.31 \times 10^{17} Hz

All Questions in this Chapter

Related Learning Resources & Exam Preparation