Question 42: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

To solve this problem, we need to understand the relationship between the atomic number $ Z $, the characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ($ \lambda_{	ext{min}} $), corresponds to the maximum energy of the X-rays produced and can be related to the accelerating voltage used in the X-ray tube. The characteristic X-rays, such as the $ K_\alpha $-line, correspond to specific electron transitions in the atom and are dependent on the atomic number $ Z $ of the target material.
The wavelength of the $ K_\alpha $-line is given by Moseley's Law, which can be approximated as:
$$ \lambda_{K_\alpha} \propto \frac{1}{(Z - 1)^2} $$
The cut-off wavelength is given by the relationship between the energy of the incident electrons and the X-ray produced:
$$ \lambda_{	ext{min}} = \frac{hc}{eV} $$
Now, given the ratio $ r $ for the first metal target with $ Z = 46 $ :
$$ r = \frac{\lambda_{K_\alpha}}{\lambda_{	ext{min}}} = 2 $$
For the second metal target with $ Z = 41 $, let's denote the new ratio as $ r_2 $. We assume the same cutoff wavelength since the same electron beam is used.
The ratio for $ Z = 41 $ can be calculated using the relationship between the $ K_\alpha $-line wavelenghts:
$$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{Z_2 - 1}{Z_1 - 1} ight)^2 $$
Substituting $ Z_1 = 46 $ and $ Z_2 = 41 $:
$$ \frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{41 - 1}{46 - 1} ight)^2 = \left( \frac{40}{45} ight)^2 = \left( \frac{8}{9} ight)^2 = \frac{64}{81} $$
Let $ \lambda_{K_\alpha (1)} = 2 \lambda_{	ext{min}} $ because $ r = 2 $. Therefore:
$$ \lambda_{K_\alpha (2)} = \lambda_{K_\alpha (1)} \cdot \frac{81}{64} = 2 \lambda_{	ext{min}} \cdot \frac{81}{64} $$
Thus, the new ratio $ r_2 $ is:
$$ r_2 = \frac{\lambda_{K_\alpha (2)}}{\lambda_{	ext{min}}} = 2 \cdot \frac{81}{64} = \frac{162}{64} = 2.53 $$
So, the value of $ r $ for the second metal target with $ Z = 41 $ is:
Option A: 2.53

Answer

2.53

Answer

1.27

Answer

2.24

Answer

1.58

Dual Nature of Matter and Radiation

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