91影视

If the frequency of ${\mathbf{K}}_{\mathbf{伪}}\mathbf{x}$ray of the spectrum of an element of an atomic number Z is v, then by Moseley鈥檚 law frequency is written as:

_{$\mathbf{谓}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{48}\mathbf{脳}{\mathbf{10}}^{\mathbf{15}}\mathbf{Hz}\mathbf{)}\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}$鈥︹赌︹赌︹赌︹赌︹赌︹赌�(1)}

The energy of the photon is determined by the equation

$\mathbf{E}\mathbf{=}\mathbf{hv}$

The relation between the wavelength and frequency of the photon is

$\mathbf{位}\mathbf{=}\frac{\mathbf{c}}{\mathbf{v}}$

Given that, the atomic number of calcium is 20.

The frequency of the${\mathrm{K}}_{\mathrm{伪}}\mathrm{x}-$ ray for the calcium:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{Ca}}=\left(2.48脳{10}^{15}\mathrm{Hz}\right){\left(20-1\right)}^2 \\ =8.95脳{10}^{17}\mathrm{Hz} \end{gathered}$$

The energy of the${\mathrm{K}}_{\mathrm{伪}}\mathrm{x}-$ ray for the calcium:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ca}}=\left(4.316脳{10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(8.95脳{10}^{17}\mathrm{Hz}\right) \\ =3.86脳{10}^3\mathrm{eV}\left(\frac{1\mathrm{keV}}{{10}^3\mathrm{eV}}\right) \\ =3.86\mathrm{keV} \end{gathered}$$

The wavelength of the$K_{伪}x-$ ray for the calcium:

$$\begin{gathered} {\mathrm{位}}_{\mathrm{Ca}}=\frac{3.0脳{10}^8\mathrm{m}/\mathrm{s}}{8.95脳{10}^{17}\mathrm{Hz}} \\ =3.35脳{10}^{-10}\mathrm{m} \end{gathered}$$

Hence, the frequency, energy, and wavelength of $K_{伪}x-$ray for the calcium are$8.95脳{10}^{17}Hz,3.86keV$ and $3.35脳{10}^{-10}\mathrm{m}$respectively.

Given that, the atomic number of cobalt is 27.

The frequency of the${\mathrm{K}}_{\mathrm{伪}}\mathrm{x}-$ray for the cobalt:

$$\begin{gathered} v_{co}=\left(2.48脳{10}^{15}Hz\right){\left(27-1\right)}^2 \\ =1.67脳{10}^{18}Hz \end{gathered}$$

The energy of the$K_{伪}x-$ray for the cobalt:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{co}}=\left(4.316脳{10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(1.68脳{10}^{18}\mathrm{Hz}\right) \\ =7.25脳{10}^3\mathrm{eV}\left(\frac{1\mathrm{keV}}{{10}^3\mathrm{eV}}\right) \\ =7.25\mathrm{keV} \end{gathered}$$

The wavelength of the$K_{伪}x-$ray for the cobalt:

$$\begin{gathered} {\mathrm{位}}_{\mathrm{Co}}=\frac{3.0脳{10}^8\mathrm{m}/\mathrm{s}}{1.68脳{10}^{18}\mathrm{Hz}} \\ =1.79脳{10}^{-10}\mathrm{m} \end{gathered}$$

Hence, the frequency, energy, and wavelength of$K_{伪}x-$ rays for the cobalt are $1.67脳{10}^{18}\mathrm{Hz},7.25鈥�\mathrm{keV}$and $1.79脳{10}^{-10}m$respectively.

Given that, the atomic number of cadmium is 48.

The frequency of the$K_{伪x-}$ ray for the cadmium:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{Cd}}=\left(2.48脳{10}^{15}\mathrm{Hz}\right){\left(48-1\right)}^2 \\ =5.48脳{10}^{18}\mathrm{Hz} \end{gathered}$$

The energy of the$K_{伪}x-$ ray for the cadmium:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cd}}=\left(4.136脳{10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(5.48脳{10}^{18}\mathrm{Hz}\right) \\ =2.36脳{10}^4\mathrm{eV}\left(\frac{1\mathrm{keV}}{{10}^3\mathrm{eV}}\right) \\ =23.6\mathrm{keV} \end{gathered}$$

The wavelength of the$K_{伪}x-$ ray for the cadmium:

$$\begin{gathered} {\mathrm{位}}_{\mathrm{Cd}}=\frac{3.0脳{10}^8\mathrm{m}/\mathrm{s}}{5.48脳{10}^{18}\mathrm{Hz}} \\ =5.48脳{10}^{-11}\mathrm{m} \end{gathered}$$

Hence, the frequency, energy, and wavelength of$K_{伪}x-$ ray for the cadmium are $5.48脳{10}^{18}\mathrm{Hz},23.6\mathrm{keV}$and $5.48脳{10}^{-11}\mathrm{m}$respectively.