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Expert-verified Found in: Page 1246 ### Fundamentals Of Physics

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" ${{\mathbit{K}}}_{{\mathbf{\alpha }}}$ line, and (c) the wavelength for the ${{\mathbit{K}}}_{{\mathbf{\beta }}}$ line increase, decrease, or remain unchanged? 1. The value of the minimum wavelength is 35.5 pm .
2. The wavelength of the ${K}_{\alpha }$ line decreases with a change in target.
3. The wavelength of the ${K}_{\beta }$ line decreases with a change in target.
See the step by step solution

## Step 1: The given data

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

## Step 2: Understanding the concept of wavelength due to transition between two states

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:

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

## Step 3: a) Calculation of the minimum wavelength for the element, molybdenum

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:

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

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

## Step 4: a) Calculation of the wavelength for the Ka line for the silver atom

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.

## Step 5: c) Calculation of the wavelength for the Kb line for the silver atom

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. ### Want to see more solutions like these? 