91Ó°ÊÓ

From figure 40-13,

The wavelength of the atomic levels in the L-shell,${\mathrm{λ}}_{\mathrm{L}}=63.0\mathrm{pm}$

The wavelength of the atomic levels in the M-shell,${\mathrm{λ}}_{\mathrm{M}}=71.0\mathrm{pm}$

Consider the known data below.

The Plank’s constant is,

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

The speed of light is,

$$\begin{gathered} c=3×{10}^8\mathrm{m}/\mathrm{s} \\ =\left(3×{10}^8×{10}^{-19}×6.63×{10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384×{10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

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

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, we can get the required energy differences between the two atomic levels by substituting the given data from Figures 40-13. Again, comparing it to the value obtained in Figure 40.15, we can get that the energy difference lies between the experiment ranges.

Formula:

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

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

Here, is the energy of the photon, is the Plank’s constant, and is the speed of light.

Rearrange equation (1) for L-shell and M-shell individually, the energy difference${\mathrm{E}}_{\mathrm{L}}-{\mathrm{E}}_{\mathrm{M}}$ for molybdenum is as follow.

${\mathrm{E}}_{\mathrm{L}}-{\mathrm{E}}_{\mathrm{M}}=\frac{\mathrm{hc}}{{\mathrm{λ}}_{\mathrm{L}}}-\frac{\mathrm{hc}}{{\mathrm{λ}}_{\mathrm{M}}}$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{L}}-{\mathrm{E}}_{\mathrm{M}}=\frac{\left(41.384×{10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3×{10}^{20}\mathrm{pm}/\mathrm{s}\right)}{63\mathrm{m}}-\frac{\left(41.384×{10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3×{10}^{20}\mathrm{pm}/\mathrm{s}\right)}{71\mathrm{m}} \\ =19.7-17.5 \\ =2.2\mathrm{keV} \end{gathered}$$

Thus, compare this value with the graphical value of figure 40-15.

The energy difference${\mathrm{E}}_{\mathrm{L}}-{\mathrm{E}}_{\mathrm{M}}$ calculated is near to the experimental value is,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{L}}-{\mathrm{E}}_{\mathrm{M}}=2.7\mathrm{keV}-0.9\mathrm{keV} \\ =1.8\mathrm{keV} \end{gathered}$$

Hence, the value of the energy difference is 2.2 keV .