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 $$K_\alpha$$-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

The electron beam has a fixed accelerating potential $$V$$, so its maximum photon energy (minimum wavelength) is the same for every target:

$$\lambda_{\text{min}} = \frac{hc}{eV} \qquad\text{(cut-off wavelength)}$$

Hence, for any two targets bombarded by the same beam, $$\lambda_{\text{min}}$$ is constant.

For the characteristic $$K_\alpha$$ X-ray emitted by a target of atomic number $$Z$$, Moseley’s law gives the frequency

$$\nu_{K_\alpha} = R \left(Z - 1\right)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}\,R\,(Z-1)^2$$

where $$R$$ is the Rydberg constant and the screening constant for the K-series is taken as 1. Since $$\nu = c/\lambda$$, the wavelength of the $$K_\alpha$$ line is

$$\lambda_{K_\alpha} = \frac{4c}{3R\,(Z-1)^2} \;\;\propto\; \frac{1}{(Z-1)^2}$$

Define $$r = \dfrac{\lambda_{K_\alpha}}{\lambda_{\text{min}}}$$. As $$\lambda_{\text{min}}$$ is the same for both targets,

$$r \;\propto\; \frac{1}{(Z-1)^2}$$

Let $$r_1$$ correspond to $$Z_1 = 46$$ and $$r_2$$ to $$Z_2 = 41$$.

$$\frac{r_2}{r_1} = \frac{1/(Z_2-1)^2}{1/(Z_1-1)^2} = \left(\frac{Z_1-1}{Z_2-1}\right)^2$$

Substitute the values:

$$(Z_1-1) = 46 - 1 = 45,\qquad (Z_2-1) = 41 - 1 = 40$$

$$\frac{r_2}{r_1} = \left(\frac{45}{40}\right)^2 = (1.125)^2 = 1.265625$$

Given $$r_1 = 2$$,

$$r_2 = 2 \times 1.265625 = 2.53125 \approx 2.53$$

Therefore, when the electron beam bombards the metal with $$Z = 41$$, the ratio $$r$$ is approximately $$2.53$$.

Option A which is: 2.53