Question

A metal target with atomic number 𝑍 = 46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio 𝑟 of the wavelengths of the 𝐾𝛼-line and the cut-off is found to be 𝑟 = 2. If the same electron beam bombards another metal target with 𝑍 = 41, the value of 𝑟 will be

• A. 2.53
• B. 1.27
• C. 2.24
• D. 1.58

Answer 1

The Correct Option is A

Calculation of the Ratio \( r \) for the Kα-Series 

For the Kα-series, the wavelength \( \lambda \) is related to the atomic number \( Z \) using the formula:

\[ \frac{1}{\lambda} = R \left( Z - 1 \right)^2 \left( 1 - \frac{1}{4} \right) \] where \( R \) is the Rydberg constant. Simplifying this expression: \[ \frac{1}{\lambda} = \frac{3}{4} R (Z - 1)^2 \]

Now, for \( Z = 46 \), the ratio \( r \) is given as:

\[ r = 2 \quad \Rightarrow \quad r = \frac{(Z - 1)^2}{(Z - 6)^2} \times 2 \] Substituting \( Z = 46 \): \[ r = \frac{(46 - 1)^2}{(46 - 6)^2} \times 2 = \frac{45^2}{40^2} \times 2 \] Simplifying: \[ r = \frac{2025}{1600} \times 2 = 2.53 \]

Final Answer:

• The ratio \( r \) is approximately: \( r \approx 2.53 \)