$\text{A metal target with atomic number }Z=46\text{ is bombarded with a high energy electron beam.}$

$\text{The emission of X-rays from the target is analyzed. }\text{The ratio}\mathrm{ r}\text{ of the wavelengths of the}{\mathrm{ K}}_{\alpha }\text{-line and}$

$\text{the cut-off is found to be}\mathrm{ r}=2.\text{If the same electron beam bombards another metal target with}\mathrm{ Z}=41,$

$\text{the value of}\mathrm{ r}\text{ will be \_\_\_\_\_}$

The correct answer is 2.53.

Step 1: Formula for Cut-off Wavelength
The minimum or cut-off wavelength (${\lambda }_{\text{min}}$) of continuous X-rays produced by an electron beam accelerating through potential difference $V$ is given by Duane-Hunt law:

${\lambda }_{\text{min}}=\frac{hc}{eV}$

Since the same electron beam is used to bombard both metal targets, the accelerating potential $V$ remains unchanged. Consequently, the cut-off wavelength ${\lambda }_{\text{min}}$ is identical for both targets.

Step 2: Formula for Characteristic $K_{\alpha }$-line Wavelength
According to Moseley's Law for characteristic X-ray emission, the wavelength ${\lambda }_{K_{\alpha }}$ corresponding to the $K_{\alpha }$ transition is given by:

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

Rearranging for ${\lambda }_{K_{\alpha }}$:

${\lambda }_{K_{\alpha }}=\frac{4}{3R{(Z-1)}^2}$

where $R$ is the Rydberg constant and $Z$ is the atomic number of the metal target.

Step 3: Finding Cut-off Wavelength using First Target ($Z_1=46$)
For the first target metal with $Z_1=46$:

${\lambda }_{K_{\alpha },1}=\frac{4}{3R{(46-1)}^2}=\frac{4}{3R{\left(45\right)}^2}=\frac{4}{3R\times 2025}=\frac{4}{6075R}$

We are given that the ratio $r_1=\frac{{\lambda }_{K_{\alpha },1}}{{\lambda }_{\text{min}}}=2$. Thus:

${\lambda }_{\text{min}}=\frac{{\lambda }_{K_{\alpha },1}}{2}=\frac{4}{2\times 6075R}=\frac{2}{6075R}$

Step 4: Calculating Ratio $r$ for Second Target ($Z_2=41$)
For the second metal target with $Z_2=41$:

${\lambda }_{K_{\alpha },2}=\frac{4}{3R{(41-1)}^2}=\frac{4}{3R{\left(40\right)}^2}=\frac{4}{3R\times 1600}=\frac{1}{1200R}$

Now, substitute ${\lambda }_{K_{\alpha },2}$ and ${\lambda }_{\text{min}}$ to find the new ratio $r_2$:

$r_2=\frac{{\lambda }_{K_{\alpha },2}}{{\lambda }_{\text{min}}}=\frac{1/\left(1200R\right)}{2/\left(6075R\right)}=\frac{6075}{2\times 1200}=\frac{6075}{2400}=2.53125$

Rounding to two decimal places, the value of $r$ for the second target is 2.53.