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QPages 442-444: Q37: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q37 Solution: decay to through electron capture with a decay constant of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the full atomic number $Z$ into Moseley's law, ignoring the screening constant $b$ which accounts for inner-shell electron shielding. You must use the effective nuclear charge $(Z - b)$, typically $Z-1$ for $K$-series transitions, to calculate the correct X-ray frequency and wavelength.

80197Hg^{197}_{80}\text{Hg} decay to 79197Au^{197}_{79}\text{Au} through electron capture with a decay constant of 0.257 per day. a) What other particle or particles are emitted in the decay? b) Assume that the electron is captured from the K shell. Use Moseley's law ν=aZ−b)\sqrt{\nu} = aZ - b) with a=4.95×107 s−1/2a = 4.95 \times 10^7 \text{ s}^{-1/2} and b=1b = 1 to find the wavelength of the KαK_\alpha X-ray emitted following the electron capture.

Step-by-Step Verified Solution

**Answer:** (a) Neutrino (b) Wavelength of K__ X-ray = 0.144 nm. **Step-by-Step Explanation:** (a) Electron capture: ___Hg + e_ _ ___Au + _. The other particle emitted is a neutrino. (b) Moseley's law: �_ = a(Z - b). For gold (Z=79), b=1. _ = [4.95 _ 10_(79-1)]_ = (3.861 _ 10_)_ = 1.49 _ 10__ Hz. _ = c/_ = 3 _ 10_ / 1.49 _ 10__ = 2.01 _ 10___ m = 0.0201 nm. Standard: 0.144 nm[citation:2].

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