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Expert-verified Found in: Page 1248 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Fig. 40-13, the x-rays shown are produced when 35.0 keV electrons strike a molybdenum (Z = 42) target. If the accelerating potential is maintained at this value but a silver (Z = 47) target is used instead, what values of (a)${{\mathbf{\lambda }}}_{{\mathbf{min}}}$ , (b) the wavelength of the ${{\mathbf{K}}}_{{\mathbf{\alpha }}}$ line, and (c) the wavelength of the ${{\mathbf{K}}}_{{\mathbf{\beta }}}$ line result? The K,L and M atomic x-ray levels for silver (compare Fig. 40-15) are 25.51, 3.56 and 0.53 keV.

1. The value of the minimum wavelength is 35.4 pm.
2. The wavelength of the ${\mathrm{K}}_{\mathrm{\alpha }}$ line is 56.5 pm.
3. The wavelength of the ${\mathrm{K}}_{\mathrm{\beta }}$ line result is .
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 target (Z = 47) is used.
3. The K,L and M levels of silver are 25.51 ,3.56 and 0.53 keV .

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

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using the basic Planck's relation, 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, the energy difference for each transition, and thus using this energy, the minimum wavelength in each case.

Formulas:

The kinetic energy gained by the electron is,

$∆\mathrm{E}=\mathrm{eV}$ ….. (1)

Here, e is the charge and V is the accelerating potential difference.

The energy of the photon due to Planck’s relation is,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

Here, h is the Plank’s constant , c is the speed of light, and $\lambda$ is the wavelength.

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

Consider the known data as below.

The Plank’s constant, $\mathrm{h}=6.63×{10}^{-34}\mathrm{J}.\mathrm{s}$

The speed of light, $\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}$

The charge, $\mathrm{e}=1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}$

Using the given data in equation (1), the value of the minimum wavelength produced by the element, molybdenum after the strike of the rays is as follow.

$\mathrm{\lambda }=\frac{\mathrm{hc}}{∆\mathrm{E}}\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{eV}}$ ….. (3)

Substitute known values in the above equation.

$\mathrm{\lambda }=\frac{6.63×{10}^{-34}×3×{10}^{8}}{35×{10}^{3}×1.6×{10}^{-19}}\phantom{\rule{0ex}{0ex}}=3.54×{10}^{-2}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=35.4\mathrm{pm}$

Hence, the value of the minimum wavelength is 35.4 pm.

## Step 4: (b) Calculation of the wavelength for the Kα line for the silver atom:

As ${K}_{\alpha }$ photon results when an electron in a target atom jumps from the L-shell to the K-shell, thus the energy of this photon is given by:

$∆\mathrm{E}=25.51\mathrm{keV}-3.56\mathrm{ke}\mathrm{V}\phantom{\rule{0ex}{0ex}}=21.95\mathrm{keV}$

Now using this value in equation (3) of part (a), the wavelength of this line is as follow.

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{eV}}\phantom{\rule{0ex}{0ex}}=\frac{6.63×{10}^{-34}×3×{10}^{8}}{21.95×{10}^{3}1.6×{10}^{-19}}\phantom{\rule{0ex}{0ex}}=5.65×{10}^{-2}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=56.5\mathrm{pm}$

Hence, the value of the wavelength produced is 56.5 pm.

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

As ${K}_{\beta }$ photon results when an electron in a target atom jumps from the M-shell to the K-shell, thus the energy of this photon is given by,

$∆\mathrm{E}=25.51\mathrm{keV}-0.53\mathrm{keV}\phantom{\rule{0ex}{0ex}}=24.98\mathrm{keV}$

Now using this value in equation (3) of part (a), the wavelength of this line is as follow.

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{eV}}\phantom{\rule{0ex}{0ex}}=\frac{\left(6.63×{10}^{-34}\right)×\left(3×{10}^{8}\right)}{\left(24.98×{10}^{3}\right)×\left(1.6×{10}^{-19}\right)}\phantom{\rule{0ex}{0ex}}=4.96×{10}^{-2}\mathrm{nm}\phantom{\rule{0ex}{0ex}}=49.6\mathrm{pm}$

Hence, the value of the wavelength produced is 49.6 pm. ### Want to see more solutions like these? 