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

1. Given Equation for \( \lambda_{\alpha} \):
We are given the following equation for \( \lambda_{\alpha} \):

$ \frac{1}{\lambda_{\alpha}} = \frac{3}{4} R (Z - 1)^2 p $

2. Expression for \( \lambda_{\text{cut}} \):
We also have the expression for \( \lambda_{\text{cut}} \):

$ \lambda_{\text{cut}} = \frac{hc}{eV} $

3. Ratio for Same Beam:
For the same beam, the ratio is:

$ \text{Ratio} \propto \frac{1}{(Z - 1)^2} $

4. Finding the Value of \( Z \) and \( x \):
We are given the relationship between \( Z \) and \( x \):

$ \frac{Z}{x} = \frac{40^2}{45^2} \Rightarrow x = \frac{45^2}{40^2} \times 2.53 \approx 2.53 $

5. Final Calculation:
Using the given values and simplifying:

$ x \approx 2.53 $

Answer 2

To solve the problem, we analyze the X-ray emission characteristics and the ratio \(r\) of wavelengths for two metal targets.

1. Background:
The cut-off wavelength \(\lambda_{\min}\) of X-rays produced by bombarding electrons of energy \(E = eV\) is given by:
\[ \lambda_{\min} = \frac{hc}{eV} \] It depends only on the energy of the electron beam and is independent of the target atomic number \(Z\).

2. Wavelength of \(K_{\alpha}\) line:
The \(K_{\alpha}\) X-ray line is due to the transition of an electron from the \(L\)-shell (n=2) to the \(K\)-shell (n=1). Its wavelength is approximately given by Moseley’s law:
\[ \frac{1}{\lambda_{K_{\alpha}}} = R (Z - 1)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R (Z - 1)^2 \frac{3}{4} \] where \(R\) is the Rydberg constant.

Rearranged:

\[ \lambda_{K_{\alpha}} = \frac{4}{3 R (Z - 1)^2} \]

3. Ratio \(r\) for the first target (Z = 46):
Given:
\[ r = \frac{\lambda_{K_{\alpha}}}{\lambda_{\min}} = 2 \] Since \(\lambda_{\min}\) depends only on the electron energy (constant for both targets), write:
\[ \lambda_{\min} = \frac{\lambda_{K_{\alpha}}}{r} = \frac{4}{3 R (46 - 1)^2 \times 2} = \frac{2}{3 R \times 45^2} \]

4. Calculate ratio \(r\) for second target \(Z=41\):
\[ \lambda_{K_{\alpha}}^{(2)} = \frac{4}{3 R (41 - 1)^2} = \frac{4}{3 R \times 40^2} \] \[ r_2 = \frac{\lambda_{K_{\alpha}}^{(2)}}{\lambda_{\min}} = \frac{\frac{4}{3 R \times 40^2}}{\frac{2}{3 R \times 45^2}} = \frac{4}{3 R \times 40^2} \times \frac{3 R \times 45^2}{2} = \frac{4 \times 45^2}{2 \times 40^2} = 2 \times \frac{45^2}{40^2} \] Calculate numerical value: \[ r_2 = 2 \times \left(\frac{45}{40}\right)^2 = 2 \times \left(\frac{9}{8}\right)^2 = 2 \times \frac{81}{64} = \frac{162}{64} = 2.53 \]

Final Answer:
\[ \boxed{2.53} \]