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

To solve this problem, we need to analyze the emission of X-rays from a metal target when bombarded with a high-energy electron beam. The key concepts involved are the Kα-line and the cut-off wavelength of the X-ray spectrum.

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Key Concepts:

1. Kα-line: This is an X-ray spectral line that arises when an electron transitions from the L-shell ($n=2$) to the K-shell ($n=1$), filling a vacancy in the K-shell. The wavelength of the Kα-line is given by Moseley's law:

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

   where:

   • $R$ is the Rydberg constant,

   • $Z$ is the atomic number of the target material,

   • $(Z-1)$ accounts for the shielding effect of the remaining electron in the K-shell.

   Simplifying, the wavelength of the Kα-line is:

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

2. Cut-off wavelength: This is the minimum wavelength of the X-ray spectrum, corresponding to the maximum energy of the incident electrons. It is given by:

   $${\lambda }_{\text{cut-off}}=\frac{hc}{E}$$

   where:

   • $h$ is Planck's constant,

   • $c$ is the speed of light,

   • $E$ is the energy of the incident electron beam.

   The cut-off wavelength is independent of the target material and depends only on the energy of the electron beam.

3. Ratio $r$: The ratio of the wavelengths of the Kα-line and the cut-off is:

   $$r=\frac{{\lambda }_{K\alpha }}{{\lambda }_{\text{cut-off}}}$$

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Step 1: Analyze the first target ($Z=46$)

For the first target ($Z=46$), the ratio $r$ is given as $r=2$. Using the expressions for ${\lambda }_{K\alpha }$ and ${\lambda }_{\text{cut-off}}$:

$$r=\frac{{\lambda }_{K\alpha }}{{\lambda }_{\text{cut-off}}}=\frac{\frac{4}{3R(Z-1{)}^2}}{{\lambda }_{\text{cut-off}}}=2$$

This implies:

$$\frac{4}{3R(Z-1{)}^2}=2{\lambda }_{\text{cut-off}}$$

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Step 2: Analyze the second target ($Z=41$)

For the second target ($Z=41$), the ratio $r$ will change because the Kα-line wavelength depends on $Z$. Using the same formula for ${\lambda }_{K\alpha }$:

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

Substituting $Z=41$:

$${\lambda }_{K\alpha }=\frac{4}{3R(41-1{)}^2}=\frac{4}{3R(40{)}^2}$$

The cut-off wavelength ${\lambda }_{\text{cut-off}}$ remains the same because it depends only on the energy of the electron beam, not on the target material.

Thus, the new ratio $r^{\mathrm{\prime }}$ is:

$$r^{\mathrm{\prime }}=\frac{{\lambda }_{K\alpha }}{{\lambda }_{\text{cut-off}}}=\frac{\frac{4}{3R(40{)}^2}}{{\lambda }_{\text{cut-off}}}$$

From Step 1, we know:

$$\frac{4}{3R(45{)}^2}=2{\lambda }_{\text{cut-off}}$$

Substituting this into the expression for $r^{\mathrm{\prime }}$:

$$r^{\mathrm{\prime }}=\frac{\frac{4}{3R(40{)}^2}}{\frac{4}{3R(45{)}^2}\cdot \frac{1}{2}}=\frac{(45{)}^2}{(40{)}^2}\cdot 2$$

Simplifying:

$$r^{\mathrm{\prime }}=2\cdot {\left(\frac{45}{40}\right)}^2=2\cdot {\left(\frac{9}{8}\right)}^2=2\cdot \frac{81}{64}=\frac{162}{64}=2.53$$

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Final Answer:

The value of $r$ for the second target ($Z=41$) is:

$$\boxed{{\displaystyle \text{(A) }2.53}}$$