From Eq. 40-26, what is the ratio of the photon energies due to transitions in two atoms whose atomic numbers are Z and Z' ? (b) What is this ratio for uranium and aluminum? (c) For uranium and lithium?

1. Atomic number of uranium, z = 92
2. Atomic number of aluminum, z = 13
3. Atomic number of lithium, z = 3

Moseley's law states that "the square root of the frequency of an atom's emitted x-ray is proportionate to its atomic number."

Formula:

According to the Moseley’s law, we can get the relation of frequency or wavelength to atomic number as:

${\mathrm{f}}_{\infty }{\left(\mathrm{Z}-1\right)}^2$OR localid="1661499422618" $\infty (\mathrm{Z}-1{)}^2$ ….. (1)

The photon energy due to Planck’s relation,

E = hf ….. (2)

From equation (2), the energy of a photon is directly proportional to the frequency of the line ${\mathrm{K}}_{\mathrm{\alpha }}$.

That means,$\mathrm{E}\infty \mathrm{f}$

So, from the above relation and the frequency equation (1), the energy is related to the atomic number of the atoms as:

$\mathrm{E}\infty {(\mathrm{Z}-1)}^2$

Now, let E and E' be the photon energies of the atoms with atomic numbers and respectively.

$\frac{\mathrm{E}}{\overline{\mathrm{E}\text{'}}}=\frac{{(\mathrm{Z}-1)}^2}{{(\mathrm{Z}\text{'}-1)}^2}$ ….. (3)

Thus, the ratio of the photon energies can be given using the above relation as follows:

Hence, the value of the required ratio is$\frac{{(\mathrm{Z}-1)}^2}{{(\mathrm{Z}\text{'}-1)}^2}$ .

Using the values of the atomic numbers of uranium and aluminum in equation (a), the ratio of the photon energies for uranium and aluminum is as follow.

$\frac{\mathrm{E}}{\mathrm{E}\text{'}}=\frac{{\left(92-1\right)}^2}{{(13-1)}^2}$

$$\begin{gathered} =\frac{{91}^2}{{12}^2} \\ =57.5 \end{gathered}$$

Hence, the value of the ratio is 57.5.

Using the values of the atomic numbers of uranium and lithium in equation (a), we can get the ratio of the photon energies for uranium and lithium as follows:

$$\begin{gathered} \frac{\mathrm{E}}{\mathrm{E}\text{'}}=\frac{{\left(92-1\right)}^2}{{\left(3-1\right)}^2} \\ =\frac{{91}^2}{2^2} \\ =2.07\times {10}^3 \end{gathered}$$

Hence, the value of the ratio is$2.07\times {10}^3$ .