HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q14 Solution: The X-rays of aluminium ) and zinc ) have wavelengths 887 pm…

QPages 395-396: Q14: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q14 Solution: The X-rays of aluminium ) and zinc ) have wavelengths 887 pm…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently insert the wavelength directly into $sqrt{\nu}=aZ-b$ instead of first converting to frequency ($\nu=c/\lambda$); this yields a linear relation with the wrong variable and leads to large errors. Always compute $\nu$ from the given $\lambda$ (using consistent SI units) before applying the square‑root law.

The

K_\alpha

X-rays of aluminium

Z = 13

) and zinc

Z = 30

) have wavelengths 887 pm and 146 pm respectively. Use Moseley's law

\sqrt{\nu} = aZ - b)

to find the wavelengths of the

K_\alpha

X-ray of iron

Z = 26

). Use Planck constant

h = 6.63 \times 10^{-34} \text{ Js} = 4.14 \times 10^{-15} \text{ eVs}

, speed of light

c = 3 \times 10^8 \text{ m/s}

Step-by-Step Verified Solution

**Answer:** Wavelength of K_alpha X-ray of iron = 193 pm. **Step-by-Step Explanation:** Given: Al (Z=13): _ = 887 pm; Zn (Z=30): _ = 146 pm. Moseley's law: �(c/_) = a(Z - b). For Al: �(3_10_/887_10___) = a(13 - b) _ 5.815 _ 10_ = a(13 - b). For Zn: �(3_10_/146_10___) = a(30 - b) _ 1.4331 _ 10_ = a(30 - b). Solving: b = 1.39, a = 5 _ 10_. For Fe (Z=26): �(c/_) = 5_10_(26 - 1.39) = 1.23_10_. _ = c/(1.23_10_)_ = 3_10_/(1.51_10__) = 1.98_10___ m = 198 pm [citation:12].

QPages 395-396: Q14: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q14 Solution: The X-rays of aluminium ) and zinc ) have wavelengths 887 pm…

Step-by-Step Verified Solution

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