1. Given data

• First target: $Z_1 = 46$
• Second target: $Z_2 = 41$
• For $Z_1=46$, the ratio $r_1 = \frac{\lambda_{K_\alpha}}{\lambda_{\min}} = 2$

We need to find the new ratio $r_2$ for the same electron beam when the target is changed to $Z_2=41$.

2. Use the cut-off wavelength relation

For an X-ray tube, $\lambda_{\min} = \frac{hc}{eV}$ Since the same electron beam is used, the accelerating voltage $V$ is unchanged.

Therefore, $\lambda_{\min}$ remains the same for both targets.

So the ratio $r$ changes only because $\lambda_{K_\alpha}$ changes.

3. Wavelength of the $K_\alpha$ line

From Moseley's law, $\nu_{K_\alpha} \propto (Z-b)^2$ For $K_\alpha$ line, the screening constant is approximately $b=1$.

Hence, $\nu_{K_\alpha} \propto (Z-1)^2$ Since $\lambda = \frac{c}{\nu},$ we get $\lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}$

Thus, $r = \frac{\lambda_{K_\alpha}}{\lambda_{\min}} \propto \lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}$

So, $\frac{r_2}{r_1} = \frac{(Z_1-1)^2}{(Z_2-1)^2}$

4. Substitute values

$r_2 = r_1\cdot \frac{(46-1)^2}{(41-1)^2}$ $r_2 = 2\cdot \frac{45^2}{40^2}$ $r_2 = 2\cdot \frac{2025}{1600}$ $r_2 = 2\cdot 1.265625$ $r_2 = 2.53125$

Therefore, $r_2 \approx 2.53$

5. Check options

• A: $2.53$ ✅
• B: $1.27$
• C: $2.24$
• D: $1.58$

So the correct option is A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.