Ratio of Wavelengths of Ka Line and Cutoff in X Ray Emission

To find the value of the ratio $r$ for the second metal target, we analyze the factors that determine both the cut-off wavelength and the $K_\alpha$-line wavelength.

1. Cut-off Wavelength ($\lambda_{\text{min}}$): The minimum wavelength (cut-off wavelength) of continuous X-rays depends solely on the accelerating potential $V$ of the incident electron beam: $\lambda_{\text{min}} = \frac{hc}{eV}$ Since the same high-energy electron beam is used for both metal targets, the accelerating voltage $V$ is identical in both cases. Therefore, the cut-off wavelength $\lambda_{\text{min}}$ remains constant for both targets.

2. Wavelength of $K_\alpha$-line ($\lambda_{K_\alpha}$): According to Moseley's Law, the wavelength of the characteristic $K_\alpha$-line for an element with atomic number $Z$ is given by: $\frac{1}{\lambda_{K_\alpha}} = R (Z - 1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4} R (Z - 1)^2$ $\implies \lambda_{K_\alpha} = \frac{4}{3R(Z - 1)^2}$ where $R$ is the Rydberg constant and the screening constant for the $K$-series is $b = 1$.

3. Ratio $r$: The ratio $r$ of the wavelength of the $K_\alpha$-line to the cut-off wavelength is: $r = \frac{\lambda_{K_\alpha}}{\lambda_{\text{min}}} = \frac{4}{3 R \lambda_{\text{min}} (Z - 1)^2}$

We can group the terms independent of $Z$ into a single constant $C = \frac{4}{3 R \lambda_{\text{min}}}$: $r = \frac{C}{(Z - 1)^2}$

1. Calculation for $Z_1 = 46$ and $Z_2 = 41$: For the first target ($Z_1 = 46$), we are given $r_1 = 2$: $r_1 = \frac{C}{(46 - 1)^2} = \frac{C}{45^2} = 2 \implies C = 2 \times 45^2$

For the second target ($Z_2 = 41$), the ratio $r_2$ is: $r_2 = \frac{C}{(41 - 1)^2} = \frac{C}{40^2}$

Substituting the expression for $C$: $r_2 = \frac{2 \times 45^2}{40^2} = 2 \times \left( \frac{45}{40} \right)^2 = 2 \times \left( \frac{9}{8} \right)^2 = 2 \times \frac{81}{64} = \frac{81}{32} \approx 2.53$

Thus, the correct value of $r$ is 2.53.

Correct Answer: (A) 2.53