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Q14Q

Expert-verifiedFound in: Page 1246

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The x-ray spectrum of Fig. 40-13 is for 35.0 keV electrons striking a molybdenum ( Z = 42 ) target. If you substitute a silver ( Z = 47 ) target for the molybdenum target, will **

**(a) ${{\mathbf{\lambda}}}_{{\mathbf{min}}}$, **

**(b) the wavelength for the role="math" localid="1661495146456" ${{\mathit{K}}}_{{\mathbf{\alpha}}}$ line, and **

**(c) the wavelength for the ${{\mathit{K}}}_{{\mathbf{\beta}}}$ line increase, decrease, or remain unchanged?**

- The value of the minimum wavelength is 35.5 pm .
- The wavelength of the ${K}_{\alpha}$ line decreases with a change in target.
- The wavelength of the ${K}_{\beta}$ line decreases with a change in target.

- The x-rays are produced due to the striking of 35 keV electrons on a molybdenum ( Z = 42 ) target.
- In the second case, a silver ( Z = 47 ) target is used.

**Using the basic Planck's relation, we can get the minimum wavelength produced due to the strike and the resulting production of the X-rays. Now, for the given emitted energy for each level for the silver atom, we can get the energy difference for each transition. Thus, using this energy, we can get the minimum wavelength in each case.**

Formulae:

The energy of the photon, due to Planck's relation:

$E=\frac{hc}{\lambda},\mathrm{where}h=6.63\times {10}^{-34}\mathrm{J}\xb7\mathrm{s},\mathrm{c}=3\times {10}^{8}\mathrm{m}/\mathrm{s}$ (i)

According to the Moseley's law, we can get the relation of frequency or wavelength to atomic number as:

$f\propto {\left(\mathrm{Z}-1\right)}^{2}\mathrm{or}\frac{\mathrm{c}}{\mathrm{\lambda}}\propto {\left(\mathrm{Z}-1\right)}^{2}$ (ii)

Using the given data in equation (i), we can get the value of the minimum wavelength produced by the element, molybdenum after the strike of the rays as follows:

$\lambda =\frac{hc}{\u2206E}\phantom{\rule{0ex}{0ex}}=\frac{\left(6.63\times {10}^{-34}\mathrm{J}\xb7\mathrm{s}\right)\left(3\times {10}^{8}\mathrm{m}/\mathrm{s}\right)}{\left(35\times {10}^{3}\mathrm{eV}\right)\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}\phantom{\rule{0ex}{0ex}}=3.55\times {10}^{-2}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=35.5\mathrm{pm}$

Hence, the value of the minimum wavelength is $35.5\mathrm{pm}$ .

Using the Moseley relation, the wavelength of a ${k}_{\alpha}$ line can be given as:

$\lambda \propto \frac{1}{{\left(Z-1\right)}^{2}}$

From the above relation, the wavelength of molybdenum can be given as:

${\lambda}_{Mb}\propto \frac{1}{\left(42-1\right)}\mathrm{or}\frac{1}{{41}^{2}}$

From the above relation, the wavelength of silver can be given as:

${\lambda}_{Ag}\propto \frac{1}{\left(47-1\right)}\mathrm{or}\frac{1}{{46}^{2}}$

Hence, the wavelength of ${K}_{\alpha}$ line referring to the atomic number of the silver in comparison to the old target, molybdenum decreases.

Similarly, considering the calculations from part (b), we can get that the wavelength of ${K}_{\beta}$ line referring to the atomic number of the silver in comparison to the old target, molybdenum decreases.

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