91影视

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

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

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

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鈥檚 relation,

$鈭�\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{位}}$ 鈥�.. (1)

Consider the known data below.

The Plank鈥檚 constant is,

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

The speed of light is,

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

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

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

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

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

$$\begin{gathered} 鈭�\mathrm{E}=\frac{1240\mathrm{keV}.\mathrm{pm}}{70\mathrm{pm}} \\ =17.7\mathrm{keV} \\ 鈮�18\mathrm{keV} \end{gathered}$$

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

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

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

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

Hence, the best suited one isZr.

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.