Question

A molybdenum (Z = 42 ) target is bombarded with 35.0keV electrons and the x-ray spectrum of Fig. 40-13 results. The lines ${\mathbf{K}}_{\mathbf{\beta }}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{K}}_{\mathbf{\alpha }}$wavelengths are 63.0 and 71.0pm, respectively. What photon energy corresponds to the (a) ${\mathbf{K}}_{\mathbf{\beta }}\mathbf{}\mathbf{and}\mathbf{}$(b) ${\mathbf{K}}_{\mathbf{\alpha }}$radiation? The two radiations are to be filtered through one of the substances in the following table such that the substance absorbs the ${\mathbf{K}}_{\mathbf{\beta }}\mathbf{}$ line more strongly than the${\mathbf{K}}_{\mathbf{\alpha }}$ line. A substance will absorb radiation ${\mathbf{x}}_{\mathbf{1}}$ more strongly than it absorbs radiation${\mathbf{x}}_{\mathbf{2}}$ if a photon of ${\mathbf{x}}_{\mathbf{1}}$ has enough energy to eject an electron ${\mathbf{K}}_{\mathbf{eiectron}}$ from an atom of the substance but a photon of does not. The table gives the ionization energy of the Kelectron in molybdenum and four other substances. Which substance in the table will serve (c) best and (d) second best as the filter?

Short answer

a) The photon energy that corresponds to the ${\mathrm{K}}_{\mathrm{\alpha }}$radiation is.20 keV .

b) The photon energy that corresponds to the ${\mathrm{K}}_{\mathrm{\beta }}$radiation is 18 keV.

c) The substance that will serve best in the filter among all the given elements in the table is Zr .

d) The substance that will serve second best in the filter among all the given elements in the table is Nb.

Step-by-step solution

The given data:

a) Wavelength of the${\mathrm{K}}_{\mathrm{\alpha }}\mathrm{line},{\mathrm{\lambda }}_{{\mathrm{K}}_{\mathrm{\alpha }}}=63\mathrm{pm}$

b) Wavelength of the${\mathrm{K}}_{\mathrm{\beta }}\mathrm{line},{\mathrm{\lambda }}_{{\mathrm{K}}_{\mathrm{\beta }}}=70\mathrm{pm}$

d) A substance absorbs ${\mathrm{x}}_1$radiation more strongly than radiation ${\mathrm{x}}_2$given that the elements absorb${\mathrm{K}}_{\mathrm{\beta }}$radiation more than the${\mathrm{K}}_{\mathrm{\alpha }}$radiation.

Understanding the concept of wavelength and radiations:

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 Planck's relation and the given wavelengths of the K-lines, to get the required energy for excitation. Again, by comparing the calculated energy for the case of required energy with elements that the element can radiate, to get the elements suitable for the experiment.

Formulae:

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

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3\times {10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

(a) Calculation of the wavelength of Kα radiation:

Using the given wavelength of the ${\mathrm{K}}_{\mathrm{\alpha }}$line in equation (1), the value of the photon energy that corresponds to the ${\mathrm{K}}_{\mathrm{\alpha }}$radiation is as follows:

$$\begin{gathered} ∆\mathrm{E}=\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{63\mathrm{pm}} \\ =\frac{1240\mathrm{keV}.\mathrm{pm}}{63\mathrm{pm}} \\ =19.7\mathrm{keV} \\ \approx 20\mathrm{keV} \end{gathered}$$

Hence, the value of the photon energy is 20 keV.

(b) Calculation of the wavelength of Kβ radiation:

Using the given wavelength of the${\mathrm{K}}_{\mathrm{\beta }}$line in equation (1), the value of the photon energy that corresponds to the${\mathrm{K}}_{\mathrm{\beta }}$radiation is as follows:

$$\begin{gathered} ∆\mathrm{E}=\frac{1240\mathrm{keV}.\mathrm{pm}}{70\mathrm{pm}} \\ =17.7\mathrm{keV} \\ \approx 18\mathrm{keV} \end{gathered}$$

Hence, the value of the photon energy is 18 keV .

(c) Calculation for the element that is best suited for this radiation:

According to the problem, the value of $K_{\beta }$radiation energy 18 keV is highly absorbed by the elements. Thus, both Zr and Nb are best suited for the usage since,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{\alpha }}<18.00\mathrm{keV}<{\mathrm{E}}_{\mathrm{\beta }},\mathrm{for}\mathrm{Zr} \\ {\mathrm{E}}_{\mathrm{\alpha }}<18.99\mathrm{keV}<{\mathrm{E}}_{\mathrm{\beta }},\mathrm{for}\mathrm{Nb} \end{gathered}$$

But among these, the best suited one is Zr as it is nearest to the photon energy value.

Hence, the best suited one isZr.

(d) Calculation for the element that is second best suited for this radiation

From the comparison in part (c), the element next to the nearest value of the photon energy is Nb .

Hence, the material second best is Nb.