Question

The wavelength of the \(K_\alpha\) line for an element of atomic number 43 is \(\lambda\). Then the wavelength of the \(K_\alpha\) line for an element of atomic number 29 is

• A. \(\left(\frac{43}{29}\right)^2 \lambda\)
• B. \(\left(\frac{42}{28}\right)^2 \lambda\)
• C. \(\left(\frac{9}{4}\right) \lambda\)
• D. \(\left(\frac{4}{9}\right) \lambda\)

Hint:

Moseley's law: \(\sqrt{\nu} = a(Z - b)\). For \(K_\alpha\), \(b = 1\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Moseley's law for \(K_\alpha\) line: \(\sqrt{\frac{1}{\lambda}} \propto (Z - 1)\) or \(\frac{1}{\lambda} \propto (Z - 1)^2\).
Step 2: Detailed Explanation:
\(\frac{1}{\lambda} = k(Z - 1)^2\).
For \(Z_1 = 43\): \(\frac{1}{\lambda} = k(42)^2\).
For \(Z_2 = 29\): \(\frac{1}{\lambda_2} = k(28)^2\).
\(\frac{\lambda_2}{\lambda} = \frac{(42)^2}{(28)^2} = \left(\frac{42}{28}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}\).
Thus, \(\lambda_2 = \frac{9}{4}\lambda\).
Step 3: Final Answer:
Thus, wavelength = \(\frac{9}{4}\lambda\).