Question

Estimate characteristic X-ray wavelengths: A hole has already been produced in the n=1 shell, and an n=2 electron is poised to jump in. Assume that the electron behaves as though in a "hydrogenlike" atom (see Section 7.8), with energy given by${\mathit{Z}}_{\mathbf{e}\mathbf{f}}^{\mathbf{2}}\left(-13.6\text{eV}/n^2\right)$ . Before the jump, it orbits Z protons, one remaining $\mathit{n}\mathbf{=}\mathbf{1}$electron. and (on average) half its seven fellow n=2 electrons, for a $Z_{\text{eff}}$of$\mathit{Z}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{5}$ . After the jump, it orbits Z protons and half of its fellow$n=1$ electron, for a $Z_{\text{eff}}$of$\mathit{Z}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{5}$ . Obtain a formula for$\mathbf{1}\mathbf{/}\mathit{\lambda }$ versus Z . Compare the predictions of this model with the experimental data given in Figure$\mathbf{8}\mathbf{.}\mathbf{19}$ and Table .

Short answer

The formula for $1/\lambda$ versus Z is $\frac{1}{\lambda }=\left(\frac{3.4}{hc}\right)\left(3Z^2+5Z-19.25\right)$ and on comparison, the theoretical and experimental values are reasonably close.

Step-by-step solution

Given data

Before the jump n=2, the electron effectively orbits an effective atomic number of$Z_e=Z-4.5$, where Z is the number of protons.

After the jump $n=1$, the electron effectively orbits an effective atomic number of $Z_e=Z=0.5$.

Concept of Energy

For a hydrogen like atom, the energy $E_n$is given by

$E_n=-13.6\text{eV}\frac{Z_e^2}{n^2}$

Where is the principal quantum number.

In terms of the wavelength of the photon , the energy difference before and after the jump$\Delta E$ is given by

$\Delta E=\frac{hc}{\lambda }$

Where his Planck's constant and is the speed of light in vacuum.

Determine the energy

When$n=2$, then $Z_2=Z-4.5$and when $n=1$then$Z_1=Z-0.5$.

The expression for $\frac{1}{\lambda }$is calculated as follows:

$$\begin{gathered} \frac{hc}{\lambda }=13.6\left[{\left(\frac{Z_1}{n_1}\right)}^2-{\left(\frac{Z_2}{n_2}\right)}^2\right] \\ \frac{hc}{\lambda }=13.6\left[{\left(\frac{Z-0.5}{1}\right)}^2-{\left(\frac{Z-4.5}{2}\right)}^2\right] \\ \frac{hc}{\lambda }=\frac{13.6}{4}\left[4\left(Z^2-Z+0.25\right)-\left(Z^2-9Z+20.25\right)\right] \\ \frac{hc}{\lambda }=3.4\left(4Z^2-4Z+1-Z^2+9Z-20.25\right) \end{gathered}$$

Simplify further as shown below.

$$\begin{gathered} \frac{hc}{\lambda }=3.4\left(3Z^2+5Z-19.25\right) \\ \frac{1}{\lambda }=\left(\frac{3.4}{hc}\right)\left(3Z^2+5Z-19.25\right) \\ \frac{1}{\lambda }=\left(\frac{3.4}{1240}\right)\left(3Z^2+5Z-19.25\right) \\ \frac{1}{\lambda }=\left(2.74\times {10}^{-3}\right)\left(3Z^2+5Z-19.25\right) \end{gathered}$$

On comparison of the calculated values with the theoretical values:

It is visible from Table (1), the theoretical value and the experimental values are reasonably close.